OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (G)

S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(g)

Question 1. \( \text{Lim}_{x \to 0} \frac{\sin 2x}{x} \)
Answer: We need to make the expression look like \( \frac{\sin \theta}{\theta} \) so we can use the standard limit, where \( \theta \) approaches 0. To do this, we multiply the denominator by 2 and also multiply the whole expression by 2 to keep it balanced.
\[ \text{Lt}_{x \to 0} \frac{\sin 2x}{x} = \text{Lt}_{x \to 0} \frac{\sin 2x}{2x} \times 2 \]Now, as \( x \to 0 \), it means \( 2x \to 0 \). We know that \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = 2 \text{Lt}_{2x \to 0} \frac{\sin 2x}{2x} = 2 \times 1 = 2 \]This special limit is a fundamental concept in calculus and helps evaluate many complex functions.
In simple words: We changed the expression so that it matches a known rule: when `sin(angle)` is divided by `angle`, and the `angle` gets very, very small (close to zero), the answer is always 1. We adjusted the numbers to make it fit this rule, and then we got 2.

🎯 Exam Tip: Remember to multiply and divide by the necessary constant to make the argument of the sine function match its denominator. This is a common trick for such limit problems.

Question 2. \( \text{Lim}_{x \to 0} \frac{\tan 2x}{x} \)
Answer: To evaluate this limit, we aim to transform the expression into the standard form \( \text{Lt}_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \). We achieve this by multiplying the denominator by 2 and compensating by multiplying the entire term by 2.
\[ \text{Lt}_{x \to 0} \frac{\tan 2x}{x} = \text{Lt}_{x \to 0} \frac{\tan 2x}{2x} \times 2 \]As \( x \to 0 \), it follows that \( 2x \to 0 \). Using the standard limit, we get:
\[ = 2 \text{Lt}_{2x \to 0} \frac{\tan 2x}{2x} = 2 \times 1 = 2 \]The relationship between sine and tangent limits is often explored in advanced trigonometry.
In simple words: We made the expression look like `tan(angle)` divided by `angle`, where `angle` goes to zero, which we know equals 1. By multiplying and dividing by 2, we got the form we needed, and the final answer is 2.

🎯 Exam Tip: Similar to the sine limit, the tangent limit also requires the argument of the tangent function to match its denominator. Always look for ways to create this structure.

Question 3. \( \text{Lim}_{x \to 0} \frac{\tan \frac{x}{2}}{3x} \)
Answer: To solve this limit, we need to arrange the expression to use the fundamental limit \( \text{Lt}_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \). Here, our \( \theta \) is \( \frac{x}{2} \).
\[ \text{Lt}_{x \to 0} \frac{\tan \frac{x}{2}}{3x} = \text{Lt}_{x \to 0} \frac{\tan \frac{x}{2}}{3x \times \frac{2}{2}} = \text{Lt}_{x \to 0} \frac{\tan \frac{x}{2}}{6 \left( \frac{x}{2} \right)} \]We can separate the constant \( \frac{1}{6} \) from the limit:
\[ = \frac{1}{6} \text{Lt}_{\frac{x}{2} \to 0} \frac{\tan \frac{x}{2}}{\frac{x}{2}} \]Since \( x \to 0 \), then \( \frac{x}{2} \to 0 \). So, the limit part becomes 1.
\[ = \frac{1}{6} \times 1 = \frac{1}{6} \]This demonstrates how constant factors can be taken outside the limit calculation.
In simple words: We wanted the bottom part to be `x/2` to match the `tan(x/2)` on top. We adjusted the `3x` to `6 * (x/2)` and then pulled the `1/6` out. Since `tan(x/2)` over `(x/2)` becomes 1 when `x` is very small, the answer is `1/6` times 1.

🎯 Exam Tip: When the argument of the trigonometric function is a fraction or a multiple of x, ensure the denominator correctly matches this argument. Constant coefficients can always be factored out of the limit operation.

Question 4. \( \text{Lim}_{x \to 0} \frac{\sin^2 5x}{\sin 15x} \)
Answer: To evaluate this limit, we will use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) multiple times. We need to introduce the appropriate denominators for each sine term.
\[ \text{Lt}_{x \to 0} \frac{\sin^2 5x}{\sin 15x} = \text{Lt}_{x \to 0} \frac{\frac{\sin^2 5x}{(5x)^2} \times (5x)^2}{\frac{\sin 15x}{15x} \times 15x} \]Now we can rearrange and simplify:
\[ = \text{Lt}_{x \to 0} \frac{\left( \frac{\sin 5x}{5x} \right)^2 \times 25x^2}{\left( \frac{\sin 15x}{15x} \right) \times 15x} \]As \( x \to 0 \), both \( 5x \to 0 \) and \( 15x \to 0 \). So, \( \left( \frac{\sin 5x}{5x} \right) \to 1 \) and \( \left( \frac{\sin 15x}{15x} \right) \to 1 \).
\[ = \frac{1^2 \times 25x^2}{1 \times 15x} = \frac{25x^2}{15x} = \frac{5x}{3} \]Now, substitute \( x = 0 \) into the simplified expression:
\[ = \frac{5 \times 0}{3} = 0 \]This problem showcases the importance of correctly applying fundamental limits to each component of a complex fraction.
In simple words: We used the `sin(angle)/angle` rule for both the top and bottom parts. We added and removed numbers like `5x` and `15x` to make the parts fit the rule. After that, we cancelled out common terms, and when we put `x=0`, the final answer became 0.

🎯 Exam Tip: For expressions involving powers of sine, make sure to apply the `(sin θ / θ)` form correctly, squaring the entire term if `sin² θ` is present. Be careful with algebraic simplification of the remaining terms.

Question 5. \( \text{Lim}_{x \to 0} \frac{\sin ax}{\sin bx} \)
Answer: To find this limit, we will use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) for both the numerator and the denominator. We introduce `ax` in the denominator for the numerator and `bx` in the denominator for the denominator.
\[ \text{Lt}_{x \to 0} \frac{\sin ax}{\sin bx} = \text{Lt}_{x \to 0} \frac{\frac{\sin ax}{ax} \times ax}{\frac{\sin bx}{bx} \times bx} \]We can rearrange the terms:
\[ = \frac{\text{Lt}_{x \to 0} \left( \frac{\sin ax}{ax} \right) \times ax}{\text{Lt}_{x \to 0} \left( \frac{\sin bx}{bx} \right) \times bx} \]As \( x \to 0 \), both \( ax \to 0 \) and \( bx \to 0 \). So, \( \frac{\sin ax}{ax} \to 1 \) and \( \frac{\sin bx}{bx} \to 1 \).
\[ = \frac{1 \times ax}{1 \times bx} = \frac{ax}{bx} = \frac{a}{b} \]This result is a useful shortcut for similar limit problems. The constants `a` and `b` determine the ratio.
In simple words: We used the `sin(angle)/angle` rule for both the top (`sin ax`) and the bottom (`sin bx`) parts. We added and removed `ax` and `bx` to make them fit the rule. After the `sin(angle)/angle` parts became 1, we were left with `ax/bx`, which simplifies to `a/b`.

🎯 Exam Tip: For limits of the form `sin(ax) / sin(bx)`, the answer is simply `a/b` when x approaches 0. However, always show the steps of multiplying and dividing by `ax` and `bx` to justify this result in exams.

Question 6. \( \text{Lim}_{x \to 0} \frac{\sin^2 ax}{\sin^2 bx} \)
Answer: To find this limit, we will apply the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). We need to create the appropriate \( \theta \) in the denominator for both the numerator and the denominator.
\[ \text{Lt}_{x \to 0} \frac{\sin^2 ax}{\sin^2 bx} = \text{Lt}_{x \to 0} \frac{\frac{\sin^2 ax}{(ax)^2} \times (ax)^2}{\frac{\sin^2 bx}{(bx)^2} \times (bx)^2} \]Rearranging the terms, we get:
\[ = \text{Lt}_{x \to 0} \frac{\left( \frac{\sin ax}{ax} \right)^2 \times a^2x^2}{\left( \frac{\sin bx}{bx} \right)^2 \times b^2x^2} \]As \( x \to 0 \), both \( ax \to 0 \) and \( bx \to 0 \). So, \( \left( \frac{\sin ax}{ax} \right) \to 1 \) and \( \left( \frac{\sin bx}{bx} \right) \to 1 \).
\[ = \frac{1^2 \times a^2x^2}{1^2 \times b^2x^2} = \frac{a^2x^2}{b^2x^2} \]We can cancel out the \( x^2 \) terms, provided \( x \neq 0 \), which is true for a limit as \( x \to 0 \).
\[ = \frac{a^2}{b^2} \]This problem illustrates how the square of the limit rule applies when the trigonometric function itself is squared.
In simple words: We used the `sin(angle)/angle` rule, but since it was `sin²`, we made `(sin(angle)/angle)²` for both top and bottom. We added `(ax)²` and `(bx)²` to make this happen. When the `sin(angle)/angle` parts became 1, we were left with `(ax)² / (bx)²`, which simplifies to `a²/b²`.

🎯 Exam Tip: Remember the property `Lt [f(x)]^n = [Lt f(x)]^n`. This allows you to apply the standard limit first and then raise the result to the power. Be careful to apply the squares to the `ax` and `bx` terms as well.

Question 7. \( \text{Lim}_{x \to 0} \frac{\sin^2 3x}{x^2} \)
Answer: To solve this limit, we need to transform the expression to utilize the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). Here, \( \theta \) is \( 3x \). We need \( (3x)^2 \) in the denominator.
\[ \text{Lt}_{x \to 0} \frac{\sin^2 3x}{x^2} = \text{Lt}_{x \to 0} \frac{\sin^2 3x}{x^2} \times \frac{9}{9} \]This step is done to get `9x²` in the denominator:
\[ = 9 \text{Lt}_{x \to 0} \frac{\sin^2 3x}{9x^2} = 9 \text{Lt}_{x \to 0} \left( \frac{\sin 3x}{3x} \right)^2 \]As \( x \to 0 \), then \( 3x \to 0 \). So, \( \frac{\sin 3x}{3x} \to 1 \).
\[ = 9 \times 1^2 = 9 \times 1 = 9 \]This problem demonstrates the importance of balancing the expression when introducing new terms into the denominator.
In simple words: We wanted the bottom part to be `(3x)²` to match `sin²(3x)` on top. We multiplied `x²` by 9 (and the whole thing by 9) to make this happen. Then, since `sin(3x)` over `3x` becomes 1, the answer is `9` times `1²`, which is `9`.

🎯 Exam Tip: When dealing with `sin²(ax)/x²`, remember to multiply and divide by `a²` to get `(sin ax / ax)²`. This ensures the argument matches its squared denominator before applying the limit.

Question 8. \( \text{Lim}_{x \to 0} \frac{\tan ax}{\tan bx} \)
Answer: To find this limit, we will use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\tan \theta}{\theta} = 1 \) for both the numerator and the denominator. We introduce `ax` in the denominator for the numerator and `bx` in the denominator for the denominator.
\[ \text{Lt}_{x \to 0} \frac{\tan ax}{\tan bx} = \text{Lt}_{x \to 0} \frac{\frac{\tan ax}{ax} \times ax}{\frac{\tan bx}{bx} \times bx} \]Rearranging the terms, we get:
\[ = \frac{\text{Lt}_{x \to 0} \left( \frac{\tan ax}{ax} \right) \times ax}{\text{Lt}_{x \to 0} \left( \frac{\tan bx}{bx} \right) \times bx} \]As \( x \to 0 \), both \( ax \to 0 \) and \( bx \to 0 \). So, \( \frac{\tan ax}{ax} \to 1 \) and \( \frac{\tan bx}{bx} \to 1 \).
\[ = \frac{1 \times ax}{1 \times bx} = \frac{ax}{bx} = \frac{a}{b} \]This result is identical to the sine version and is very common in limit calculations. It is useful for quickly solving problems.
In simple words: Just like with `sin`, we used the `tan(angle)/angle` rule for both the top (`tan ax`) and the bottom (`tan bx`) parts. After the `tan(angle)/angle` parts became 1, we were left with `ax/bx`, which simplifies to `a/b`.

🎯 Exam Tip: The limit `Lt (tan ax / tan bx)` as `x->0` is `a/b`. This is a direct consequence of the fundamental `tan θ / θ` limit. Always show the intermediate steps of multiplying and dividing by `ax` and `bx` to demonstrate your understanding.

Question 9. \( \text{Lim}_{x \to 0} \frac{\sin^2 x}{2x} \)
Answer: To evaluate this limit, we need to manipulate the expression to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ \text{Lt}_{x \to 0} \frac{\sin^2 x}{2x} = \text{Lt}_{x \to 0} \frac{1}{2} \times \frac{\sin^2 x}{x} \]We can rewrite \( \frac{\sin^2 x}{x} \) as \( \frac{\sin^2 x}{x^2} \times x \).
\[ = \frac{1}{2} \text{Lt}_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \times x \]As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \).
\[ = \frac{1}{2} \times 1^2 \times 0 = \frac{1}{2} \times 0 = 0 \]Even though part of the expression goes to 1, the `x` term drives the overall limit to zero. Understanding which terms become zero and which become one is crucial.
In simple words: We separated `sin²x/2x` into `(1/2) * (sin²x / x)`. Then, we made `sin²x / x` look like `(sin x / x)² * x`. Since `sin x / x` becomes 1, the expression became `(1/2) * 1² * x`. When `x` goes to 0, the whole thing becomes `(1/2) * 1 * 0`, which is 0.

🎯 Exam Tip: Be careful when an extra `x` term remains after applying the `sin θ / θ` limit. If that `x` term goes to zero, the entire expression will typically go to zero, unless it's in the denominator.

Question 10. \( \text{Lim}_{x \to 0} \frac{\sin x^2}{x} \)
Answer: To evaluate this limit, we need to make the expression conform to the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). Here, our \( \theta \) is \( x^2 \). We need \( x^2 \) in the denominator.
\[ \text{Lt}_{x \to 0} \frac{\sin x^2}{x} = \text{Lt}_{x \to 0} \frac{\sin x^2}{x^2} \times x \]Now, as \( x \to 0 \), it means \( x^2 \to 0 \). So, \( \text{Lt}_{x^2 \to 0} \frac{\sin x^2}{x^2} = 1 \).
\[ = \left( \text{Lt}_{x^2 \to 0} \frac{\sin x^2}{x^2} \right) \times \left( \text{Lt}_{x \to 0} x \right) \]Substituting the limits:
\[ = 1 \times 0 = 0 \]This problem illustrates that even if the argument is a power of `x`, the fundamental limit rule still applies. Identifying the correct argument is key.
In simple words: We wanted the bottom part to be `x²` to match `sin(x²)` on top. So, we changed `x` to `x²` on the bottom and multiplied by `x` to keep it balanced. Since `sin(x²)` over `x²` becomes 1, and the extra `x` becomes 0, the answer is `1 * 0`, which is 0.

🎯 Exam Tip: When the argument of the sine function is `x²`, make sure the denominator is also `x²`. Any remaining `x` terms after applying the fundamental limit rule will determine if the overall limit is zero or another value.

Question 11. \( \text{Lim}_{x \to 0} \frac{\sin^3 ax}{\sin^2 bx} \)
Answer: To evaluate this limit, we use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) for both the numerator and the denominator.
\[ \text{Lt}_{x \to 0} \frac{\sin^3 ax}{\sin^2 bx} = \text{Lt}_{x \to 0} \frac{\frac{\sin^3 ax}{(ax)^3} \times (ax)^3}{\frac{\sin^2 bx}{(bx)^2} \times (bx)^2} \]Rearrange and apply the limit rule:
\[ = \text{Lt}_{x \to 0} \frac{\left( \frac{\sin ax}{ax} \right)^3 \times a^3x^3}{\left( \frac{\sin bx}{bx} \right)^2 \times b^2x^2} \]As \( x \to 0 \), \( \left( \frac{\sin ax}{ax} \right) \to 1 \) and \( \left( \frac{\sin bx}{bx} \right) \to 1 \).
\[ = \frac{1^3 \times a^3x^3}{1^2 \times b^2x^2} = \frac{a^3x^3}{b^2x^2} \]Simplify the `x` terms:
\[ = \frac{a^3}{b^2} x \]Now, substitute \( x = 0 \) into the simplified expression:
\[ = \frac{a^3}{b^2} \times 0 = 0 \]The powers of `x` in the numerator and denominator play a crucial role in determining the final limit value.
In simple words: We used the `sin(angle)/angle` rule for both the top (`sin³ ax`) and bottom (`sin² bx`) parts. We adjusted them by adding `(ax)³` and `(bx)²` to make the rule work. After the `sin(angle)/angle` parts became 1, we were left with `(a³x³) / (b²x²)`, which simplifies to `(a³/b²) * x`. When `x` goes to 0, the answer becomes 0.

🎯 Exam Tip: When dealing with higher powers of sine, make sure to adjust the denominators by multiplying and dividing by the corresponding power of the argument (e.g., `(ax)³` for `sin³ ax`). Simplify the resulting `x` terms carefully before taking the final limit.

Question 12. \( \text{Lim}_{x \to 0} \frac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} \)
Answer: We will use the sum-to-product trigonometric formulas to simplify the numerator and denominator: * \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) \) * \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \) Applying these to the expression:
Numerator: \( \sin 2x + \sin 6x = 2 \sin\left(\frac{2x+6x}{2}\right) \cos\left(\frac{2x-6x}{2}\right) = 2 \sin(4x) \cos(-2x) = 2 \sin(4x) \cos(2x) \) (since \( \cos(-\theta) = \cos \theta \))
Denominator: \( \sin 5x - \sin 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \sin\left(\frac{5x-3x}{2}\right) = 2 \cos(4x) \sin(x) \)
So the expression becomes:
\[ \text{Lt}_{x \to 0} \frac{2 \sin 4x \cos 2x}{2 \cos 4x \sin x} = \text{Lt}_{x \to 0} \frac{\sin 4x \cos 2x}{\cos 4x \sin x} \]Now, we can separate the terms and use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = \text{Lt}_{x \to 0} \frac{\frac{\sin 4x}{4x} \times 4x \times \cos 2x}{\cos 4x \times \frac{\sin x}{x} \times x} \]Rearrange terms:
\[ = \frac{\text{Lt}_{x \to 0} \frac{\sin 4x}{4x} \times \text{Lt}_{x \to 0} \cos 2x \times \text{Lt}_{x \to 0} 4x}{\text{Lt}_{x \to 0} \cos 4x \times \text{Lt}_{x \to 0} \frac{\sin x}{x} \times \text{Lt}_{x \to 0} x} \]As \( x \to 0 \), \( \frac{\sin 4x}{4x} \to 1 \), \( \frac{\sin x}{x} \to 1 \), \( \cos 2x \to \cos 0 = 1 \), and \( \cos 4x \to \cos 0 = 1 \).
\[ = \frac{1 \times 1 \times \text{Lt}_{x \to 0} 4x}{1 \times 1 \times \text{Lt}_{x \to 0} x} = \frac{4x}{x} = 4 \]Trigonometric identities are powerful tools for simplifying expressions before applying limits. The limit of `cos(ax)` as `x` approaches 0 is always 1.
In simple words: First, we used special math rules to change the `sin + sin` on top and `sin - sin` on the bottom into `2 sin cos` and `2 cos sin`. Then, we cancelled out the 2s. Next, we split the expression and used the `sin(angle)/angle` rule for `sin 4x` and `sin x`. When `x` gets very small, `cos 2x` and `cos 4x` both become 1. After all the cancelling and using rules, we were left with `4x/x`, which simplifies to 4.

🎯 Exam Tip: Always look for opportunities to apply sum-to-product or product-to-sum trigonometric identities to simplify expressions before applying limits, especially when dealing with sums or differences of sine/cosine terms.

Question 13. \( \text{Lim}_{x \to 0} \frac{\cos mx - \cos nx}{x^2} \)
Answer: We use the trigonometric identity for \( \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Applying this to the numerator:
\[ \cos mx - \cos nx = -2 \sin\left(\frac{mx+nx}{2}\right) \sin\left(\frac{mx-nx}{2}\right) \]\[ = -2 \sin\left(\frac{(m+n)x}{2}\right) \sin\left(\frac{(m-n)x}{2}\right) \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{-2 \sin\left(\frac{(m+n)x}{2}\right) \sin\left(\frac{(m-n)x}{2}\right)}{x^2} \]Now, we need to introduce the correct denominators for each sine term to apply \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = -2 \text{Lt}_{x \to 0} \frac{\sin\left(\frac{(m+n)x}{2}\right)}{\frac{(m+n)x}{2}} \times \frac{(m+n)x}{2} \times \frac{\sin\left(\frac{(m-n)x}{2}\right)}{\frac{(m-n)x}{2}} \times \frac{(m-n)x}{2} \times \frac{1}{x^2} \]Rearranging and simplifying:
\[ = -2 \text{Lt}_{x \to 0} \left( \frac{\sin\left(\frac{(m+n)x}{2}\right)}{\frac{(m+n)x}{2}} \right) \times \left( \frac{\sin\left(\frac{(m-n)x}{2}\right)}{\frac{(m-n)x}{2}} \right) \times \frac{(m+n)x}{2} \times \frac{(m-n)x}{2} \times \frac{1}{x^2} \]As \( x \to 0 \), both \( \frac{(m+n)x}{2} \to 0 \) and \( \frac{(m-n)x}{2} \to 0 \). The sine limit parts become 1.
\[ = -2 \times 1 \times 1 \times \frac{(m+n)(m-n)x^2}{4x^2} \]Cancel \( x^2 \) and simplify the constants:
\[ = -2 \times \frac{(m^2 - n^2)}{4} = -\frac{(m^2 - n^2)}{2} = \frac{n^2 - m^2}{2} \]This is a standard result for this type of limit and can be very useful. The ability to correctly apply trig identities is a powerful skill.
In simple words: We used a special rule to change `cos - cos` into `-2 sin sin`. Then, we adjusted the numbers under each `sin` part to make them match `sin(angle)/angle`, which becomes 1. After all the `sin` parts became 1, we were left with `-2 * (m+n)x/2 * (m-n)x/2` divided by `x²`. This simplified to `(n² - m²) / 2`.

🎯 Exam Tip: The identity `cos C - cos D = -2 sin((C+D)/2) sin((C-D)/2)` is crucial for these types of limits. Remember to properly manage the `x²` in the denominator by using the arguments of the sine functions to form `(sin θ / θ)` terms.

Question 14. \( \text{Lim}_{x \to 0} \frac{2 \sin^2 3x}{x^2} \)
Answer: To evaluate this limit, we aim to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). Here, our \( \theta \) is \( 3x \), so we need \( (3x)^2 \) or \( 9x^2 \) in the denominator.
\[ \text{Lt}_{x \to 0} \frac{2 \sin^2 3x}{x^2} = 2 \text{Lt}_{x \to 0} \frac{\sin^2 3x}{x^2} \]To get \( 9x^2 \) in the denominator, we multiply and divide by 9:
\[ = 2 \text{Lt}_{x \to 0} \frac{\sin^2 3x}{9x^2} \times 9 \]Now, we can rewrite \( \frac{\sin^2 3x}{9x^2} \) as \( \left( \frac{\sin 3x}{3x} \right)^2 \).
\[ = 2 \times 9 \text{Lt}_{x \to 0} \left( \frac{\sin 3x}{3x} \right)^2 \]As \( x \to 0 \), then \( 3x \to 0 \). So, \( \frac{\sin 3x}{3x} \to 1 \).
\[ = 18 \times 1^2 = 18 \times 1 = 18 \]This highlights how constant factors are carried through the limit calculation. This limit is often used in understanding wave behavior.
In simple words: We wanted the bottom part to be `(3x)²` to match `sin²(3x)` on top. So, we multiplied `x²` by 9 (and the whole thing by 9) to make this happen. Then, since `sin(3x)` over `3x` becomes 1, the answer is `2 * 9` times `1²`, which is 18.

🎯 Exam Tip: Remember to adjust the denominator by multiplying by the square of the coefficient of `x` (e.g., `3² = 9` for `3x`) to create the `(sin ax / ax)²` form. Don't forget to multiply the entire expression by the same factor to maintain balance.

Question 15. \( \text{Lim}_{x \to 0} \frac{1 - \cos 2x}{x} \)
Answer: We use the trigonometric identity \( 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \).
Applying this to the numerator with \( A = 2x \):
\[ 1 - \cos 2x = 2 \sin^2 \left( \frac{2x}{2} \right) = 2 \sin^2 x \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{2 \sin^2 x}{x} \]Now, we can manipulate this expression to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = \text{Lt}_{x \to 0} 2 \times \frac{\sin x}{x} \times \sin x \]As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \) and \( \sin x \to \sin 0 = 0 \).
\[ = 2 \times 1 \times 0 = 0 \]This result demonstrates that when a term approaches zero, it can cause the entire limit to be zero. Understanding trigonometric identities is key here.
In simple words: First, we used a rule to change `1 - cos 2x` into `2 sin² x`. Then, we split the expression `(2 sin² x) / x` into `2 * (sin x / x) * sin x`. Since `sin x / x` becomes 1 and `sin x` becomes 0 when `x` is very small, the final answer is `2 * 1 * 0`, which is 0.

🎯 Exam Tip: The identity `1 - cos A = 2 sin²(A/2)` is fundamental for limits involving `(1 - cos A)`. Always simplify using this identity first, then apply the standard `sin θ / θ` limit and evaluate any remaining trigonometric terms.

Question 16. \( \text{Lim}_{x \to 0} \frac{1 - \cos 4x}{x^2} \)
Answer: We use the trigonometric identity \( 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \).
Applying this to the numerator with \( A = 4x \):
\[ 1 - \cos 4x = 2 \sin^2 \left( \frac{4x}{2} \right) = 2 \sin^2 2x \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{2 \sin^2 2x}{x^2} \]Now, we need to manipulate this expression to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). We need \( (2x)^2 \) or \( 4x^2 \) in the denominator for \( \sin^2 2x \).
\[ = 2 \text{Lt}_{x \to 0} \frac{\sin^2 2x}{x^2} \times \frac{4}{4} = 2 \times 4 \text{Lt}_{x \to 0} \frac{\sin^2 2x}{4x^2} \]Rearranging into the standard form:
\[ = 8 \text{Lt}_{x \to 0} \left( \frac{\sin 2x}{2x} \right)^2 \]As \( x \to 0 \), then \( 2x \to 0 \). So, \( \frac{\sin 2x}{2x} \to 1 \).
\[ = 8 \times 1^2 = 8 \times 1 = 8 \]This result is frequently used in physics, particularly in wave optics. It highlights the symmetry between the numerator and denominator powers.
In simple words: First, we used a rule to change `1 - cos 4x` into `2 sin² 2x`. Then, to use the `sin(angle)/angle` rule, we needed `(2x)²` (which is `4x²`) under `sin² 2x`. So, we multiplied `x²` by 4 and balanced it by multiplying the whole thing by 4. After that, `(sin 2x) / (2x)` became 1, so the answer is `2 * 4` times `1²`, which is 8.

🎯 Exam Tip: For `(1 - cos ax) / x²` type problems, the answer is `a²/2`. You should show the intermediate steps by converting `1 - cos ax` to `2 sin²(ax/2)` and then adjusting the denominator to `(ax/2)²` to properly apply the limit formula.

Question 17. \( \text{Lim}_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx} \)
Answer: We use the trigonometric identity \( 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \) for both the numerator and the denominator.
Numerator: \( 1 - \cos mx = 2 \sin^2 \left( \frac{mx}{2} \right) \)
Denominator: \( 1 - \cos nx = 2 \sin^2 \left( \frac{nx}{2} \right) \)
So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{2 \sin^2 \left( \frac{mx}{2} \right)}{2 \sin^2 \left( \frac{nx}{2} \right)} = \text{Lt}_{x \to 0} \frac{\sin^2 \left( \frac{mx}{2} \right)}{\sin^2 \left( \frac{nx}{2} \right)} \]Now, we apply the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \) to both numerator and denominator. We need \( \left( \frac{mx}{2} \right)^2 \) and \( \left( \frac{nx}{2} \right)^2 \) in their respective denominators.
\[ = \text{Lt}_{x \to 0} \frac{\frac{\sin^2 \left( \frac{mx}{2} \right)}{\left( \frac{mx}{2} \right)^2} \times \left( \frac{mx}{2} \right)^2}{\frac{\sin^2 \left( \frac{nx}{2} \right)}{\left( \frac{nx}{2} \right)^2} \times \left( \frac{nx}{2} \right)^2} \]As \( x \to 0 \), then \( \frac{mx}{2} \to 0 \) and \( \frac{nx}{2} \to 0 \). So, the \( \frac{\sin \theta}{\theta} \) parts become 1.
\[ = \frac{1^2 \times \frac{m^2x^2}{4}}{1^2 \times \frac{n^2x^2}{4}} = \frac{m^2x^2}{n^2x^2} \]Cancel \( x^2 \) and \( 4 \):
\[ = \frac{m^2}{n^2} \]This problem illustrates how fundamental identities simplify complex fraction limits. This result is widely applicable in many areas of engineering.
In simple words: We used a rule to change `1 - cos` into `2 sin²` for both the top and bottom. The `2`s cancelled out. Then, for both the top and bottom, we made the expression look like `(sin(angle)/angle)²`. We added `(mx/2)²` and `(nx/2)²` to make this happen. When the `(sin(angle)/angle)²` parts became 1, we were left with `(m²x²/4) / (n²x²/4)`, which simplifies to `m²/n²`.

🎯 Exam Tip: For limits of the form `(1 - cos mx) / (1 - cos nx)`, the answer is `m²/n²`. This is a very common result. Ensure you show the steps of using the `1 - cos A` identity and then applying the `(sin θ / θ)²` form for full marks.

Question 18. \( \text{Lim}_{x \to 0} \frac{\cos Ax - \cos Bx}{x^2} \)
Answer: We use the trigonometric identity for \( \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Applying this to the numerator:
\[ \cos Ax - \cos Bx = -2 \sin\left(\frac{Ax+Bx}{2}\right) \sin\left(\frac{Ax-Bx}{2}\right) \]\[ = -2 \sin\left(\frac{(A+B)x}{2}\right) \sin\left(\frac{(A-B)x}{2}\right) \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{-2 \sin\left(\frac{(A+B)x}{2}\right) \sin\left(\frac{(A-B)x}{2}\right)}{x^2} \]Now, we introduce the correct denominators for each sine term to apply \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = -2 \text{Lt}_{x \to 0} \frac{\sin\left(\frac{(A+B)x}{2}\right)}{\frac{(A+B)x}{2}} \times \frac{(A+B)x}{2} \times \frac{\sin\left(\frac{(A-B)x}{2}\right)}{\frac{(A-B)x}{2}} \times \frac{(A-B)x}{2} \times \frac{1}{x^2} \]Rearranging and simplifying:
\[ = -2 \text{Lt}_{x \to 0} \left( \frac{\sin\left(\frac{(A+B)x}{2}\right)}{\frac{(A+B)x}{2}} \right) \times \left( \frac{\sin\left(\frac{(A-B)x}{2}\right)}{\frac{(A-B)x}{2}} \right) \times \frac{(A+B)(A-B)x^2}{4x^2} \]As \( x \to 0 \), the sine limit parts become 1.
\[ = -2 \times 1 \times 1 \times \frac{(A^2 - B^2)}{4} \]Cancel 2 from numerator and denominator:
\[ = -\frac{(A^2 - B^2)}{2} = \frac{B^2 - A^2}{2} \]This limit is a generalized form of a common derivative result. Understanding product and sum formulas is important here.
In simple words: We used the `cos C - cos D` rule to change the top part into `-2 sin sin`. Then, we adjusted the numbers under each `sin` to make them match `sin(angle)/angle`, which becomes 1. After the `sin` parts became 1, we were left with `-2 * (A+B)x/2 * (A-B)x/2` divided by `x²`. This simplified to `(B² - A²) / 2`.

🎯 Exam Tip: This limit is a direct application of the `cos C - cos D` identity. Pay close attention to the negative sign in the identity and ensure careful algebraic simplification of the constants after applying the `sin θ / θ` limit for each sine term.

Question 19. \( \text{Lim}_{x \to 0} \frac{3 \sin x - \sin 3x}{x^3} \)
Answer: We use the trigonometric identity \( \sin 3x = 3 \sin x - 4 \sin^3 x \).
Rearranging this identity to find \( 3 \sin x - \sin 3x \):
\[ 3 \sin x - \sin 3x = 3 \sin x - (3 \sin x - 4 \sin^3 x) \]\[ = 3 \sin x - 3 \sin x + 4 \sin^3 x = 4 \sin^3 x \]So, the limit expression becomes:
\[ \text{Lt}_{x \to 0} \frac{4 \sin^3 x}{x^3} \]Now, we can rewrite this to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\[ = 4 \text{Lt}_{x \to 0} \left( \frac{\sin x}{x} \right)^3 \]As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \).
\[ = 4 \times 1^3 = 4 \times 1 = 4 \]This problem is a good example of how memorizing trigonometric identities can simplify complex limit calculations. This identity is important in Fourier series.
In simple words: We used a special math rule (`sin 3x = 3 sin x - 4 sin³ x`) to change the top part of the fraction. This made the top simply `4 sin³ x`. Then, we made the whole expression look like `4` times `(sin x / x)³`. Since `sin x / x` becomes 1 when `x` is very small, the answer is `4 * 1³`, which is 4.

🎯 Exam Tip: The identity `sin 3x = 3 sin x - 4 sin³ x` is crucial for this problem. Always look for ways to simplify complex trigonometric expressions in the numerator or denominator using such identities before applying standard limit forms.

Question 20. \( \text{Lim}_{x \to 0} \frac{\sin 3x \cos 2x}{\sin 2x} \)
Answer: We need to manipulate this expression to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
We can split the expression and introduce appropriate terms:
\[ \text{Lt}_{x \to 0} \frac{\sin 3x \cos 2x}{\sin 2x} = \text{Lt}_{x \to 0} \frac{\frac{\sin 3x}{3x} \times 3x \times \cos 2x}{\frac{\sin 2x}{2x} \times 2x} \]Rearranging the terms for clarity:
\[ = \frac{\text{Lt}_{x \to 0} \frac{\sin 3x}{3x} \times \text{Lt}_{x \to 0} \cos 2x \times \text{Lt}_{x \to 0} 3x}{\text{Lt}_{x \to 0} \frac{\sin 2x}{2x} \times \text{Lt}_{x \to 0} 2x} \]As \( x \to 0 \), \( \frac{\sin 3x}{3x} \to 1 \), \( \cos 2x \to \cos 0 = 1 \), and \( \frac{\sin 2x}{2x} \to 1 \).
\[ = \frac{1 \times 1 \times 3x}{1 \times 2x} \]Cancel \( x \) from numerator and denominator:
\[ = \frac{3}{2} \]This problem demonstrates that individual components of a limit expression can be evaluated separately. This approach is helpful for breaking down complex functions.
In simple words: We split the expression and used the `sin(angle)/angle` rule for `sin 3x` and `sin 2x`. We also know that `cos 2x` becomes `cos 0`, which is 1, when `x` is very small. After applying these rules and cancelling out the `x` terms, we were left with `3/2`.

🎯 Exam Tip: When faced with a product of functions, evaluate the limits of individual functions if they exist. Here, `cos 2x` can be evaluated directly, while `sin 3x` and `sin 2x` require the `sin θ / θ` adjustment.

Question 21. \( \text{Lim}_{x \to 0} \frac{x^2}{1 - \cos x} \)
Answer: We use the trigonometric identity \( 1 - \cos A = 2 \sin^2 \left( \frac{A}{2} \right) \).
Applying this to the denominator with \( A = x \):
\[ 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right) \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{x^2}{2 \sin^2 \left( \frac{x}{2} \right)} \]Now, we need to manipulate this expression to use the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). We need \( \left( \frac{x}{2} \right)^2 \) or \( \frac{x^2}{4} \) in the denominator for \( \sin^2 \left( \frac{x}{2} \right) \).
\[ = \text{Lt}_{x \to 0} \frac{x^2}{2 \times \frac{\sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2} \times \left( \frac{x}{2} \right)^2} \]Rearranging and simplifying:
\[ = \text{Lt}_{x \to 0} \frac{x^2}{2 \times \left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)^2 \times \frac{x^2}{4}} \]As \( x \to 0 \), then \( \frac{x}{2} \to 0 \). So, \( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \to 1 \).
\[ = \frac{x^2}{2 \times 1^2 \times \frac{x^2}{4}} = \frac{x^2}{\frac{2x^2}{4}} = \frac{x^2}{\frac{x^2}{2}} \]Multiplying by the reciprocal:
\[ = x^2 \times \frac{2}{x^2} = 2 \]This problem shows that even if terms appear complex, using appropriate identities can simplify them significantly. This technique is often used in the study of oscillations.
In simple words: First, we used a rule to change `1 - cos x` into `2 sin² (x/2)`. Then, for the bottom part, we made it look like `2` times `(sin(x/2) / (x/2))²` and balanced it by adding `(x/2)²`. Since `(sin(x/2) / (x/2))` becomes 1, the bottom became `2 * 1² * (x²/4)`, which is `x²/2`. So the whole fraction was `x² / (x²/2)`, which simplifies to 2.

🎯 Exam Tip: This is a reciprocal of a common limit. The key steps are using the `1 - cos A` identity and ensuring the denominator for `sin²(A/2)` is `(A/2)²`. Careful simplification of `x²` terms will lead to the correct numerical answer.

Question 22. \( \text{Lim}_{x \to 0} \frac{\sin 3x - \sin x}{\sin x} \)
Answer: We use the trigonometric identity for \( \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) \).
Applying this to the numerator with \( C = 3x \) and \( D = x \):
\[ \sin 3x - \sin x = 2 \cos\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) \]\[ = 2 \cos\left(\frac{4x}{2}\right) \sin\left(\frac{2x}{2}\right) = 2 \cos(2x) \sin(x) \]So, the limit becomes:
\[ \text{Lt}_{x \to 0} \frac{2 \cos(2x) \sin(x)}{\sin(x)} \]We can cancel \( \sin(x) \) from the numerator and denominator, provided \( \sin(x) \neq 0 \). For a limit as \( x \to 0 \), \( x \) gets very close to 0 but is not equal to 0, so \( \sin(x) \neq 0 \).
\[ = \text{Lt}_{x \to 0} 2 \cos(2x) \]Now, substitute \( x = 0 \) into the expression:
\[ = 2 \cos(2 \times 0) = 2 \cos(0) \]Since \( \cos(0) = 1 \):
\[ = 2 \times 1 = 2 \]This simplification demonstrates how canceling common factors can greatly reduce the complexity of limit problems. The identity is very useful in signal processing.
In simple words: First, we used a special rule to change `sin 3x - sin x` on top into `2 cos 2x sin x`. Then, we noticed that `sin x` appears on both the top and the bottom, so we cancelled them out. We were left with `2 cos 2x`. When `x` gets very small (approaches 0), `cos 2x` becomes `cos 0`, which is 1. So, the final answer is `2 * 1`, which is 2.

🎯 Exam Tip: Always try to simplify trigonometric expressions using identities before applying limits. If a common factor can be canceled, it often leads to a much simpler limit evaluation, as seen with `sin x` in this problem.