Samacheer Kalvi Class 11 Maths Solutions Chapter 9 Limits and Continuity Exercise 9.2

Question 1. \( \lim_{x \to 2} \frac{x^4 - 16}{x - 2} \)
Answer:
We have the limit expression: \( \lim_{x \to 2} \frac{x^4 - 16}{x - 2} \)
We can rewrite the numerator using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \):
\( = \lim_{x \to 2} \frac{(x^2)^2 - 4^2}{x - 2} \)
\( = \lim_{x \to 2} \frac{(x^2 - 4)(x^2 + 4)}{x - 2} \)
Now, we apply the difference of squares formula again to \( x^2 - 4 \):
\( = \lim_{x \to 2} \frac{(x^2 - 2^2)(x^2 + 4)}{x - 2} \)
\( = \lim_{x \to 2} \frac{(x - 2)(x + 2)(x^2 + 4)}{x - 2} \)
Since \( x \to 2 \) but \( x \neq 2 \), we can cancel out the \( (x - 2) \) term from the numerator and the denominator:
\( = \lim_{x \to 2} (x + 2)(x^2 + 4) \)
Now, substitute \( x = 2 \) into the simplified expression:
\( = (2 + 2)(2^2 + 4) \)
\( = 4 \times (4 + 4) \)
\( = 4 \times 8 \)
\( = 32 \)
This method helps to simplify the expression and avoid division by zero when the limit is approached.
In simple words: We broke down the top part of the fraction using a special math rule. Then we cancelled out a common part from both top and bottom. After that, we put the number 2 into the simplified sum to get our final answer.

🎯 Exam Tip: When evaluating limits that result in an indeterminate form like \( \frac{0}{0} \), always look for algebraic simplifications, such as factoring or rationalizing, to remove the problematic term.

Question 2. \( \lim_{x \to 1} \frac{x^m - 1}{x^n - 1} \), m and n are integers.
Answer:
We have the limit expression: \( \lim_{x \to 1} \frac{x^m - 1}{x^n - 1} \)
We can rewrite \( 1 \) as \( 1^m \) and \( 1^n \):
\( = \lim_{x \to 1} \frac{x^m - 1^m}{x^n - 1^n} \)
To use the standard limit formula \( \lim_{x \to a} \frac{x^k - a^k}{x - a} = ka^{k-1} \), we can divide both the numerator and the denominator by \( (x - 1) \):
\( = \lim_{x \to 1} \frac{\frac{x^m - 1^m}{x - 1}}{\frac{x^n - 1^n}{x - 1}} \)
Now, we apply the limit formula to both the numerator and the denominator. Here, \( a = 1 \):
For the numerator, \( k = m \): \( m(1)^{m-1} = m \)
For the denominator, \( k = n \): \( n(1)^{n-1} = n \)
So, the limit becomes:
\( = \frac{m(1)^{m-1}}{n(1)^{n-1}} \)
Since any power of \( 1 \) is \( 1 \):
\( = \frac{m \times 1}{n \times 1} \)
\( = \frac{m}{n} \)
This formula is a very useful shortcut for solving limits of expressions that look like a power minus a constant.
In simple words: We changed the numbers a bit to use a special limit rule. Then we applied the rule to both the top and bottom parts of the fraction. After simplifying, we found the answer is simply \( m \) divided by \( n \).

🎯 Exam Tip: Remember the limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \). This is a crucial shortcut for many limit problems, especially when \(a=1\).

Question 3. \( \lim_{x \to 3} \frac{x^2 - 81}{\sqrt{x} - 3} \)
Answer:
We have the limit expression: \( \lim_{x \to 3} \frac{x^2 - 81}{\sqrt{x} - 3} \)
Let's use a substitution to simplify the expression. Let \( y = \sqrt{x} \).
This means \( x = y^2 \).
When \( x \to 3 \), then \( y \to \sqrt{3} \).
Wait, looking at the steps in the solution, it replaces \( \sqrt{x} \) with \( y \) and treats the constant \( 3 \) as part of the denominator. Let's follow the solution's steps where \( \sqrt{x} = y \), and when \( \sqrt{x} \to 3 \), it means \( y \to 3 \).
If \( \sqrt{x} = y \), then \( x = y^2 \).
Also, \( x^2 = (y^2)^2 = y^4 \).
So, the expression becomes:
\( = \lim_{y \to 3} \frac{y^4 - 81}{y - 3} \)
We can rewrite \( 81 \) as \( 3^4 \):
\( = \lim_{y \to 3} \frac{y^4 - 3^4}{y - 3} \)
Now, we apply the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = na^{n-1} \).
Here, \( n = 4 \) and \( a = 3 \):
\( = 4(3)^{4-1} \)
\( = 4 \times 3^3 \)
\( = 4 \times 27 \)
\( = 108 \)
Changing the variable can often make a complex limit problem easier to recognize and solve using standard formulas.
In simple words: We changed \( \sqrt{x} \) to \( y \) to make the sum easier to work with. Then we used a known limit rule to solve for \( y \). Finally, we put the value \( y = 3 \) into the simplified expression to get our final answer.

🎯 Exam Tip: When dealing with square roots in limit problems, substituting \( y = \sqrt{x} \) can transform the expression into a more recognizable form for applying algebraic limit formulas.

Question 4. \( \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \), \( x > 0 \)
Answer:
We have the limit expression: \( \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \)
This limit is a direct application of the definition of the derivative. To solve it using the limit formula \( \lim_{z \to a} \frac{z^n - a^n}{z - a} = na^{n-1} \), we can rearrange the terms.
Let \( z = x+h \). As \( h \to 0 \), then \( z \to x \).
The expression becomes:
\( = \lim_{z \to x} \frac{z^{\frac{1}{2}} - x^{\frac{1}{2}}}{z - x} \)
Now, apply the limit formula with \( n = \frac{1}{2} \) and \( a = x \):
\( = \frac{1}{2} (x)^{\frac{1}{2} - 1} \)
\( = \frac{1}{2} x^{-\frac{1}{2}} \)
\( = \frac{1}{2\sqrt{x}} \)
This limit is actually the definition of the derivative of \( \sqrt{x} \), which is a fundamental concept in calculus.
In simple words: We rewrote the fraction to match a common limit rule. By applying this rule, we were able to quickly find the answer which simplifies to one over two times the square root of \( x \).

🎯 Exam Tip: Recognize this specific limit as the definition of the derivative of \( \sqrt{x} \). This is a foundational result that often appears in calculus questions.

Question 5. \( \lim_{x \to 5} \frac{\sqrt{x+4} - 3}{x - 5} \)
Answer:
We have the limit expression: \( \lim_{x \to 5} \frac{\sqrt{x+4} - 3}{x - 5} \)
Let's use a substitution to simplify this problem. Let \( y = x+4 \).
\(\implies \) From this, we can say \( x = y - 4 \).
\(\implies \) So, \( x - 5 = (y - 4) - 5 = y - 9 \).
Now, consider the limit for \( y \). When \( x \to 5 \), then \( y \to 5+4 = 9 \).
Substitute these into the original limit expression:
\( = \lim_{y \to 9} \frac{\sqrt{y} - 3}{y - 9} \)
We can rewrite \( \sqrt{y} \) as \( y^{\frac{1}{2}} \) and \( 3 \) as \( 9^{\frac{1}{2}} \):
\( = \lim_{y \to 9} \frac{y^{\frac{1}{2}} - 9^{\frac{1}{2}}}{y - 9} \)
Now, we apply the standard limit formula \( \lim_{z \to a} \frac{z^n - a^n}{z - a} = na^{n-1} \).
Here, \( n = \frac{1}{2} \) and \( a = 9 \):
\( = \frac{1}{2} (9)^{\frac{1}{2} - 1} \)
\( = \frac{1}{2} (9)^{-\frac{1}{2}} \)
\( = \frac{1}{2} \times \frac{1}{\sqrt{9}} \)
\( = \frac{1}{2} \times \frac{1}{3} \)
\( = \frac{1}{6} \)
This substitution method makes the expression fit a known limit theorem, allowing for a straightforward calculation.
In simple words: We changed \( x+4 \) to \( y \) and found what \( y \) becomes as \( x \) approaches 5. This helped us use a special limit rule. We then put the values into the rule and simplified to get the final fraction as our answer.

🎯 Exam Tip: For limits involving square roots and linear expressions, try using substitution to convert them into the \( \frac{x^n - a^n}{x - a} \) form. This simplifies the evaluation significantly.

Question 6. \( \lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \)
Answer:
We have the limit expression: \( \lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \)
First, simplify the numerator by finding a common denominator for the fractions:
\( = \lim_{x \to 2} \frac{\frac{2 - x}{2x}}{x - 2} \)
Now, we can rewrite the expression by multiplying the numerator by the reciprocal of the denominator:
\( = \lim_{x \to 2} \frac{2 - x}{2x(x - 2)} \)
Notice that \( (2 - x) \) is the negative of \( (x - 2) \). So, \( (2 - x) = -(x - 2) \):
\( = \lim_{x \to 2} \frac{-(x - 2)}{2x(x - 2)} \)
Since \( x \to 2 \) but \( x \neq 2 \), we can cancel out the \( (x - 2) \) term from the numerator and the denominator:
\( = \lim_{x \to 2} \frac{-1}{2x} \)
Now, substitute \( x = 2 \) into the simplified expression:
\( = \frac{-1}{2 \times 2} \)
\( = -\frac{1}{4} \)
Direct substitution initially gives an undefined form, so algebraic manipulation is crucial to find the true limit.
In simple words: First, we simplified the top part of the fraction by finding a common bottom number. Then we took out a minus sign to make the \( (x-2) \) parts match so we could cancel them. Finally, we put \( x = 2 \) into the simplified sum to get the answer.

🎯 Exam Tip: When simplifying complex fractions for limits, always look for opportunities to factor out or rewrite terms (like \( 2-x = -(x-2) \)) to allow for cancellation and avoid indeterminate forms.

Question 7. \( \lim_{x \to 1} \frac{\sqrt{x} - x^2}{1 - \sqrt{x}} \)
Answer:
We have the limit expression: \( \lim_{x \to 1} \frac{\sqrt{x} - x^2}{1 - \sqrt{x}} \)
Let's use a substitution to simplify this problem. Let \( y = \sqrt{x} \).
\(\implies \) This means \( x = y^2 \).
\(\implies \) Also, \( x^2 = (y^2)^2 = y^4 \).
Now, consider the limit for \( y \). When \( x \to 1 \), then \( y \to \sqrt{1} = 1 \).
Substitute these into the original limit expression:
\( = \lim_{y \to 1} \frac{y - y^4}{1 - y} \)
Factor out \( y \) from the numerator:
\( = \lim_{y \to 1} \frac{y(1 - y^3)}{1 - y} \)
Now, use the difference of cubes formula, \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \), for \( (1 - y^3) \):
\( = \lim_{y \to 1} \frac{y(1 - y)(1^2 + 1 \cdot y + y^2)}{1 - y} \)
Since \( y \to 1 \) but \( y \neq 1 \), we can cancel out the \( (1 - y) \) term from the numerator and the denominator:
\( = \lim_{y \to 1} y(1 + y + y^2) \)
Now, substitute \( y = 1 \) into the simplified expression:
\( = 1(1 + 1 + 1^2) \)
\( = 1 \times (1 + 1 + 1) \)
\( = 1 \times 3 \)
\( = 3 \)
Factoring algebraic expressions is a powerful technique to resolve indeterminate forms in limits.
In simple words: We changed \( \sqrt{x} \) to \( y \) to make the sum easier. Then we simplified the fraction by taking out \( y \) and breaking down \( 1-y^3 \) using a math rule. After cancelling out a common part, we put \( y = 1 \) into the simplified expression to find the answer.

🎯 Exam Tip: Algebraic identities like difference of squares or cubes are extremely helpful in simplifying expressions within limits, especially when direct substitution leads to \( \frac{0}{0} \).

Question 8. \( \lim_{x \to 0} \frac{\sqrt{x^2+1} - 1}{\sqrt{x^2+16} - 4} \)
Answer:
We have the limit expression: \( \lim_{x \to 0} \frac{\sqrt{x^2+1} - 1}{\sqrt{x^2+16} - 4} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 0 \). To resolve this, we will rationalize both the numerator and the denominator by multiplying by their respective conjugates.
Multiply by the conjugate of the numerator, \( (\sqrt{x^2+1} + 1) \), and the conjugate of the denominator, \( (\sqrt{x^2+16} + 4) \):
\( = \lim_{x \to 0} \left( \frac{\sqrt{x^2+1} - 1}{\sqrt{x^2+16} - 4} \times \frac{\sqrt{x^2+1} + 1}{\sqrt{x^2+1} + 1} \times \frac{\sqrt{x^2+16} + 4}{\sqrt{x^2+16} + 4} \right) \)
Apply the difference of squares formula, \( (a-b)(a+b) = a^2 - b^2 \):
\( = \lim_{x \to 0} \frac{( (x^2+1) - 1^2 ) (\sqrt{x^2+16} + 4)}{( (x^2+16) - 4^2 ) (\sqrt{x^2+1} + 1)} \)
\( = \lim_{x \to 0} \frac{(x^2+1 - 1)(\sqrt{x^2+16} + 4)}{(x^2+16 - 16)(\sqrt{x^2+1} + 1)} \)
Simplify the terms:
\( = \lim_{x \to 0} \frac{x^2(\sqrt{x^2+16} + 4)}{x^2(\sqrt{x^2+1} + 1)} \)
Since \( x \to 0 \) but \( x \neq 0 \), we can cancel out the \( x^2 \) term from the numerator and the denominator:
\( = \lim_{x \to 0} \frac{\sqrt{x^2+16} + 4}{\sqrt{x^2+1} + 1} \)
Now, substitute \( x = 0 \) into the simplified expression:
\( = \frac{\sqrt{0^2+16} + 4}{\sqrt{0^2+1} + 1} \)
\( = \frac{\sqrt{16} + 4}{\sqrt{1} + 1} \)
\( = \frac{4 + 4}{1 + 1} \)
\( = \frac{8}{2} \)
\( = 4 \)
Multiplying by the conjugate is a common strategy to simplify expressions involving square roots and eliminate indeterminate forms.
In simple words: We multiplied the top and bottom by special terms called conjugates to remove the square roots. After simplifying and cancelling out \( x^2 \), we put \( x = 0 \) into the new fraction. Then we solved it to get our final answer.

🎯 Exam Tip: For limits with multiple square roots in both the numerator and denominator, remember to rationalize both by multiplying by their respective conjugates to simplify the expression effectively.

Question 9. \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \)
Answer:
We have the limit expression: \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 0 \). To resolve this, we will rationalize the numerator by multiplying by its conjugate.
Multiply both the numerator and the denominator by \( (\sqrt{1+x} + 1) \):
\( = \lim_{x \to 0} \left( \frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} \right) \)
Apply the difference of squares formula, \( (a-b)(a+b) = a^2 - b^2 \), to the numerator:
\( = \lim_{x \to 0} \frac{(1+x) - 1^2}{x(\sqrt{1+x} + 1)} \)
\( = \lim_{x \to 0} \frac{1+x - 1}{x(\sqrt{1+x} + 1)} \)
Simplify the numerator:
\( = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} \)
Since \( x \to 0 \) but \( x \neq 0 \), we can cancel out the \( x \) term from the numerator and the denominator:
\( = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \)
Now, substitute \( x = 0 \) into the simplified expression:
\( = \frac{1}{\sqrt{1+0} + 1} \)
\( = \frac{1}{\sqrt{1} + 1} \)
\( = \frac{1}{1 + 1} \)
\( = \frac{1}{2} \)
This limit is another form of the derivative definition, specifically for \( \sqrt{x} \) at \( x=1 \) shifted by 1.
In simple words: We multiplied the top and bottom by the conjugate of the numerator. This helped us simplify the expression by removing the square root and cancelling \( x \). Then we put \( x = 0 \) into the new fraction and solved it to get the answer.

🎯 Exam Tip: Rationalizing the numerator is a standard technique for limits involving square roots. Always look for this step when direct substitution yields \( \frac{0}{0} \).

Question 10. \( \lim_{x \to 1} \frac{\sqrt[3]{7+x^2} - \sqrt{3+x^2}}{x - 1} \)
Answer:
We have the limit expression: \( \lim_{x \to 1} \frac{\sqrt[3]{7+x^2} - \sqrt{3+x^2}}{x - 1} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 1 \) because \( \sqrt[3]{7+1^2} = \sqrt[3]{8} = 2 \) and \( \sqrt{3+1^2} = \sqrt{4} = 2 \), so the numerator is \( 2-2=0 \).
To solve this, we will add and subtract 2 in the numerator, which is the value of the expressions at \( x=1 \):
\( = \lim_{x \to 1} \frac{\left( (7+x^2)^{\frac{1}{3}} - 2 \right) - \left( (3+x^2)^{\frac{1}{2}} - 2 \right)}{x - 1} \)
We can split this into two separate limits:
\( = \lim_{x \to 1} \frac{(7+x^2)^{\frac{1}{3}} - 2}{x - 1} - \lim_{x \to 1} \frac{(3+x^2)^{\frac{1}{2}} - 2}{x - 1} \)
Let's evaluate each limit separately using the definition of the derivative or L'Hôpital's Rule. For a function \( f(x) \), \( \lim_{x \to a} \frac{f(x) - f(a)}{x-a} = f'(a) \).
For the first limit, let \( f_1(x) = (7+x^2)^{\frac{1}{3}} \). Then \( f_1(1) = (7+1^2)^{\frac{1}{3}} = 8^{\frac{1}{3}} = 2 \).
\( f_1'(x) = \frac{1}{3}(7+x^2)^{\frac{1}{3}-1} \times (2x) = \frac{2x}{3}(7+x^2)^{-\frac{2}{3}} \)
\( f_1'(1) = \frac{2(1)}{3}(7+1^2)^{-\frac{2}{3}} = \frac{2}{3}(8)^{-\frac{2}{3}} = \frac{2}{3} \times \frac{1}{8^{\frac{2}{3}}} = \frac{2}{3} \times \frac{1}{(\sqrt[3]{8})^2} = \frac{2}{3} \times \frac{1}{2^2} = \frac{2}{3} \times \frac{1}{4} = \frac{1}{6} \)
So, \( \lim_{x \to 1} \frac{(7+x^2)^{\frac{1}{3}} - 2}{x - 1} = \frac{1}{6} \)

For the second limit, let \( f_2(x) = (3+x^2)^{\frac{1}{2}} \). Then \( f_2(1) = (3+1^2)^{\frac{1}{2}} = 4^{\frac{1}{2}} = 2 \).
\( f_2'(x) = \frac{1}{2}(3+x^2)^{\frac{1}{2}-1} \times (2x) = x(3+x^2)^{-\frac{1}{2}} = \frac{x}{\sqrt{3+x^2}} \)
\( f_2'(1) = \frac{1}{\sqrt{3+1^2}} = \frac{1}{\sqrt{4}} = \frac{1}{2} \)
So, \( \lim_{x \to 1} \frac{(3+x^2)^{\frac{1}{2}} - 2}{x - 1} = \frac{1}{2} \)

Combining the results:
\( = \frac{1}{6} - \frac{1}{2} \)
\( = \frac{1}{6} - \frac{3}{6} \)
\( = -\frac{2}{6} \)
\( = -\frac{1}{3} \)
When solving limits with multiple roots, subtracting and adding a constant can help break the problem into simpler parts.
In simple words: We split the complex limit into two simpler parts by adding and subtracting a number. For each part, we used a special rule involving derivatives. Finally, we subtracted the second result from the first to get the total answer.

🎯 Exam Tip: For limits involving sums or differences of complex functions (like different roots), a common strategy is to add and subtract the function's value at the limit point in the numerator to apply derivative definitions or L'Hôpital's Rule effectively.

Question 11. \( \lim_{x \to 2} \frac{2 - \sqrt{x+2}}{\sqrt[3]{2} - \sqrt[3]{4-x}} \)
Answer:
We have the limit expression: \( \lim_{x \to 2} \frac{2 - \sqrt{x+2}}{\sqrt[3]{2} - \sqrt[3]{4-x}} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 2 \) because \( 2 - \sqrt{2+2} = 2 - \sqrt{4} = 0 \) and \( \sqrt[3]{2} - \sqrt[3]{4-2} = \sqrt[3]{2} - \sqrt[3]{2} = 0 \).
We can solve this using L'Hôpital's Rule since it's an indeterminate form. L'Hôpital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) is of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \).
Let \( f(x) = 2 - \sqrt{x+2} = 2 - (x+2)^{\frac{1}{2}} \).
Let \( g(x) = \sqrt[3]{2} - \sqrt[3]{4-x} = 2^{\frac{1}{3}} - (4-x)^{\frac{1}{3}} \).
Find the derivative of \( f(x) \):
\( f'(x) = 0 - \frac{1}{2}(x+2)^{\frac{1}{2}-1} \times (1) = -\frac{1}{2}(x+2)^{-\frac{1}{2}} = -\frac{1}{2\sqrt{x+2}} \)
Evaluate \( f'(x) \) at \( x = 2 \):
\( f'(2) = -\frac{1}{2\sqrt{2+2}} = -\frac{1}{2\sqrt{4}} = -\frac{1}{2 \times 2} = -\frac{1}{4} \)
Find the derivative of \( g(x) \):
\( g'(x) = 0 - \frac{1}{3}(4-x)^{\frac{1}{3}-1} \times (-1) = \frac{1}{3}(4-x)^{-\frac{2}{3}} = \frac{1}{3\sqrt[3]{(4-x)^2}} \)
Evaluate \( g'(x) \) at \( x = 2 \):
\( g'(2) = \frac{1}{3}(4-2)^{-\frac{2}{3}} = \frac{1}{3}(2)^{-\frac{2}{3}} = \frac{1}{3 \times 2^{\frac{2}{3}}} = \frac{1}{3\sqrt[3]{2^2}} = \frac{1}{3\sqrt[3]{4}} \)
Now, apply L'Hôpital's Rule:
\( \lim_{x \to 2} \frac{f'(x)}{g'(x)} = \frac{f'(2)}{g'(2)} = \frac{-\frac{1}{4}}{\frac{1}{3\sqrt[3]{4}}} \)
\( = -\frac{1}{4} \times (3\sqrt[3]{4}) \)
\( = -\frac{3\sqrt[3]{4}}{4} \)
When dealing with indeterminate forms, using L'Hôpital's Rule by taking derivatives of the numerator and denominator can often simplify the limit calculation.
In simple words: We used a special rule called L'Hôpital's rule. This means we found the derivative of the top part and the derivative of the bottom part of the fraction. Then we put \( x = 2 \) into these new parts and divided them to get our final answer.

🎯 Exam Tip: L'Hôpital's Rule is very powerful for \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate forms, but ensure both functions are differentiable and the limit of their derivatives exists.

Question 12. \( \lim_{x \to 0} \frac{\sqrt{1+x^2} - 1}{x} \)
Answer:
We have the limit expression: \( \lim_{x \to 0} \frac{\sqrt{1+x^2} - 1}{x} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 0 \). To resolve this, we will rationalize the numerator by multiplying by its conjugate.
Multiply both the numerator and the denominator by \( (\sqrt{1+x^2} + 1) \):
\( = \lim_{x \to 0} \left( \frac{\sqrt{1+x^2} - 1}{x} \times \frac{\sqrt{1+x^2} + 1}{\sqrt{1+x^2} + 1} \right) \)
Apply the difference of squares formula, \( (a-b)(a+b) = a^2 - b^2 \), to the numerator:
\( = \lim_{x \to 0} \frac{(1+x^2) - 1^2}{x(\sqrt{1+x^2} + 1)} \)
\( = \lim_{x \to 0} \frac{1+x^2 - 1}{x(\sqrt{1+x^2} + 1)} \)
Simplify the numerator:
\( = \lim_{x \to 0} \frac{x^2}{x(\sqrt{1+x^2} + 1)} \)
Since \( x \to 0 \) but \( x \neq 0 \), we can cancel out one \( x \) term from the numerator and the denominator:
\( = \lim_{x \to 0} \frac{x}{\sqrt{1+x^2} + 1} \)
Now, substitute \( x = 0 \) into the simplified expression:
\( = \frac{0}{\sqrt{1+0^2} + 1} \)
\( = \frac{0}{\sqrt{1} + 1} \)
\( = \frac{0}{1 + 1} \)
\( = \frac{0}{2} \)
\( = 0 \)
Rationalizing the numerator is a key step to eliminate the indeterminate form and evaluate the limit directly.
In simple words: We multiplied the top and bottom by the special conjugate of the top part. This helped us simplify the fraction by removing the square root and cancelling \( x \). Then we put \( x = 0 \) into the new fraction and solved it to get zero as our answer.

🎯 Exam Tip: Be careful when cancelling terms like \( x \) or \( x^2 \). Ensure you are only cancelling what is valid for \( x \neq 0 \), which is true when evaluating limits.

Question 13. \( \lim_{x \to 0} \frac{\sqrt{1-x} - 1}{x^2} \)
Answer:
We have the limit expression: \( \lim_{x \to 0} \frac{\sqrt{1-x} - 1}{x^2} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 0 \). To resolve this, we will rationalize the numerator by multiplying by its conjugate.
Multiply both the numerator and the denominator by \( (\sqrt{1-x} + 1) \):
\( = \lim_{x \to 0} \left( \frac{\sqrt{1-x} - 1}{x^2} \times \frac{\sqrt{1-x} + 1}{\sqrt{1-x} + 1} \right) \)
Apply the difference of squares formula, \( (a-b)(a+b) = a^2 - b^2 \), to the numerator:
\( = \lim_{x \to 0} \frac{(1-x) - 1^2}{x^2(\sqrt{1-x} + 1)} \)
\( = \lim_{x \to 0} \frac{1-x - 1}{x^2(\sqrt{1-x} + 1)} \)
Simplify the numerator:
\( = \lim_{x \to 0} \frac{-x}{x^2(\sqrt{1-x} + 1)} \)
Since \( x \to 0 \) but \( x \neq 0 \), we can cancel out one \( x \) term from the numerator and the denominator:
\( = \lim_{x \to 0} \frac{-1}{x(\sqrt{1-x} + 1)} \)
Now, if we substitute \( x = 0 \), the numerator is \( -1 \) and the denominator becomes \( 0 \times (\sqrt{1-0} + 1) = 0 \times (1+1) = 0 \).
Since the numerator is a non-zero constant and the denominator approaches \( 0 \), the limit does not exist.
More specifically, as \( x \to 0^+ \) (from the positive side), \( x \) is a small positive number, so \( x(\sqrt{1-x} + 1) \) is positive. Thus, \( \frac{-1}{\text{small positive number}} \to -\infty \).
As \( x \to 0^- \) (from the negative side), \( x \) is a small negative number, so \( x(\sqrt{1-x} + 1) \) is negative. Thus, \( \frac{-1}{\text{small negative number}} \to +\infty \).
Since the limit approaches different values from the left and right, the limit does not exist.
When a simplified limit results in a non-zero number divided by zero, it indicates that the limit does not exist, often approaching positive or negative infinity.
In simple words: We first multiplied by a special term to remove the square root. After simplifying the fraction, we tried to put \( x = 0 \). Since the top part was a number and the bottom part became zero, the limit does not exist. It approaches negative infinity from one side and positive infinity from the other.

🎯 Exam Tip: If, after simplification, a limit results in \( \frac{C}{0} \) (where \( C \neq 0 \)), the limit does not exist. Determine if it approaches \( \pm \infty \) by checking limits from the left and right.

Question 14. \( \lim_{x \to 5} \frac{\sqrt{x-1} - 2}{x - 5} \)
Answer:
We have the limit expression: \( \lim_{x \to 5} \frac{\sqrt{x-1} - 2}{x - 5} \)
This is an indeterminate form \( \frac{0}{0} \) when \( x = 5 \). To resolve this, let's use a substitution.
Let \( y = x - 1 \).
Now, consider the limit for \( y \). When \( x \to 5 \), then \( y \to 5 - 1 = 4 \).
Substitute these into the original limit expression:
\( = \lim_{y \to 4} \frac{\sqrt{y} - 2}{y - 4} \)
We can rewrite \( \sqrt{y} \) as \( y^{\frac{1}{2}} \) and \( 2 \) as \( 4^{\frac{1}{2}} \):
\( = \lim_{y \to 4} \frac{y^{\frac{1}{2}} - 4^{\frac{1}{2}}}{y - 4} \)
Now, we apply the standard limit formula \( \lim_{z \to a} \frac{z^n - a^n}{z - a} = na^{n-1} \).
Here, \( n = \frac{1}{2} \) and \( a = 4 \):
\( = \frac{1}{2} (4)^{\frac{1}{2} - 1} \)
\( = \frac{1}{2} (4)^{-\frac{1}{2}} \)
\( = \frac{1}{2} \times \frac{1}{\sqrt{4}} \)
\( = \frac{1}{2} \times \frac{1}{2} \)
\( = \frac{1}{4} \)
This technique of variable substitution simplifies the expression, allowing the direct application of a fundamental limit theorem.
In simple words: We replaced \( x-1 \) with \( y \) to make the limit look like a standard formula. Then, we used the limit rule \( na^{n-1} \) to quickly find the answer, which came out as one fourth.

🎯 Exam Tip: Remember that substituting variables can transform a seemingly complex limit into a standard form, making it much easier to solve using known limit theorems.

Question 15. \( \lim_{x \to a} \frac { \sqrt { x-b } - \sqrt { a-b } }{ x^{2} - a^{2} } ( a > b ) \)
Answer:
We have the limit: \( \lim_{x \to a} \frac { \sqrt { x-b } - \sqrt { a-b } }{ x^{2} - a^{2} } \)
To solve this, we multiply the numerator and denominator by the conjugate of the numerator:
\( = \lim_{x \to a} \frac { \sqrt { x-b } - \sqrt { a-b } }{ x^{2} - a^{2} } \times \frac { \sqrt { x-b } + \sqrt { a-b } }{ \sqrt { x-b } + \sqrt { a-b } } \)
Now, we use the difference of squares formula \( (A-B)(A+B) = A^2 - B^2 \) in the numerator and factor \( x^2 - a^2 \) as \( (x-a)(x+a) \) in the denominator:
\( = \lim_{x \to a} \frac { (x-b) - (a-b) }{ (x^{2} - a^{2}) [ \sqrt { x-b } + \sqrt { a-b } ] } \)
\( = \lim_{x \to a} \frac { x - b - a + b }{ (x-a)(x+a) [ \sqrt { x-b } + \sqrt { a-b } ] } \)
\( = \lim_{x \to a} \frac { x - a }{ (x-a)(x+a) [ \sqrt { x-b } + \sqrt { a-b } ] } \)
Cancel out the common term \( (x-a) \):
\( = \lim_{x \to a} \frac { 1 }{ (x+a) [ \sqrt { x-b } + \sqrt { a-b } ] } \)
Substitute \( x = a \) into the expression:
\( = \frac { 1 }{ (a+a) [ \sqrt { a-b } + \sqrt { a-b } ] } \)
\( = \frac { 1 }{ (2a) [ 2\sqrt { a-b } ] } \)
\( = \frac { 1 }{ 4a\sqrt { a-b } } \)
In simple words: To find this limit, we multiply the top and bottom by a special term called the conjugate to simplify the fraction. Then we cancel out common parts and finally put the value of 'a' in place of 'x' to get the answer.

🎯 Exam Tip: When dealing with limits involving square roots, always try to multiply by the conjugate to simplify the expression and eliminate the indeterminate form. Remember to factor algebraic expressions like \( x^2 - a^2 \) carefully.