Maharashtra Board Class 11 Maths Part 1 Chapter 7 Limits Miscellaneous Solutions

Std 11 Maths 1 Miscellaneous Exercise 7 Solutions Commerce Maths

I.

Question 1. If \( \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80 \) then find the value of n.
Answer:
Solution:
We are given \( \lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80 \)
Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \), we get:
\( n(2)^{n-1} = 80 \)
We need to express 80 in terms of powers of 2.
\( n(2)^{n-1} = 5 \times 16 \)
\( n(2)^{n-1} = 5 \times 2^4 \)
Comparing this with \( n(2)^{n-1} = 5 \times 2^{5-1} \)
Therefore, \( n = 5 \)
In simple words: We used the standard limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) to solve for 'n' by comparing the given equation with the formula and equating the corresponding parts.

🎯 Exam Tip: Remember the standard limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \) as it is frequently tested. Pay attention to simplifying numerical values into powers of the base 'a'.

II. Evaluate the Following Limits:

Question 1. Evaluate \( \lim_{x \to a} \frac{(x+2)^{5/3} - (a+2)^{5/3}}{x - a} \)
Answer:
Solution:
We have \( \lim_{x \to a} \frac{(x+2)^{5/3} - (a+2)^{5/3}}{x - a} \)
Let \( y = x+2 \) and \( b = a+2 \).
As \( x \to a \), then \( x+2 \to a+2 \), which means \( y \to b \).
The limit becomes:
\( \lim_{y \to b} \frac{y^{5/3} - b^{5/3}}{(y-2) - (b-2)} \)
\( = \lim_{y \to b} \frac{y^{5/3} - b^{5/3}}{y-b} \)
Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \),
\( = \frac{5}{3} b^{\frac{5}{3} - 1} \)
\( = \frac{5}{3} b^{\frac{2}{3}} \)
Substitute back \( b = a+2 \):
\( = \frac{5}{3} (a+2)^{\frac{2}{3}} \)
In simple words: We used substitution to transform the given limit into the standard form \( \lim_{y \to b} \frac{y^n - b^n}{y-b} \), and then applied the power rule for limits to find the result.

🎯 Exam Tip: When the limit expression is not in the standard \( \frac{x^n-a^n}{x-a} \) form, consider substitution to simplify it. Ensure all terms (the variable approaching the limit, and the base) are consistently substituted.

Question 2. Evaluate \( \lim_{x \to 0} \frac{(1+x)^n - 1}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{(1+x)^n - 1}{x} \)
Let \( y = 1+x \). Then \( x = y-1 \).
As \( x \to 0 \), then \( y \to 1 \).
The limit becomes:
\( \lim_{y \to 1} \frac{y^n - 1}{y-1} \)
Using the formula \( \lim_{y \to a} \frac{y^n - a^n}{y-a} = na^{n-1} \) (here \( a=1 \)),
\( = n(1)^{n-1} \)
\( = n \)
In simple words: We used substitution to change the variable and transform the expression into the standard limit form, where the base is 1, simplifying the calculation to 'n'.

🎯 Exam Tip: Remember that \( 1^k = 1 \) for any real number k. This simplifies calculations when 'a' in the limit formula is 1. Substitution is key for adapting expressions to standard forms.

Question 3. Evaluate \( \lim_{x \to 2} \frac{(x-2)}{2x^2-7x+6} \)
Answer:
Solution:
We have \( \lim_{x \to 2} \frac{x-2}{2x^2-7x+6} \)
First, factor the denominator \( 2x^2-7x+6 \).
By splitting the middle term: \( 2x^2 - 4x - 3x + 6 = 2x(x-2) - 3(x-2) = (x-2)(2x-3) \).
So, \( \lim_{x \to 2} \frac{x-2}{(x-2)(2x-3)} \)
Since \( x \to 2 \), \( x \neq 2 \), so \( x-2 \neq 0 \). We can cancel \( (x-2) \) from numerator and denominator.
\( = \lim_{x \to 2} \frac{1}{2x-3} \)
Substitute \( x=2 \):
\( = \frac{1}{2(2)-3} \)
\( = \frac{1}{4-3} \)
\( = 1 \)
In simple words: We factored the quadratic expression in the denominator, canceled out the common term that was causing the indeterminate form, and then directly substituted the limit value to get the answer.

🎯 Exam Tip: Always try to factorize or simplify the expression to remove any indeterminate forms (like \( \frac{0}{0} \)) before direct substitution. This often involves factoring quadratic or cubic expressions.

Question 4. Evaluate \( \lim_{x \to 1} \frac{x^3-1}{x^2+5x-6} \)
Answer:
Solution:
We have \( \lim_{x \to 1} \frac{x^3-1}{x^2+5x-6} \)
Factor the numerator using \( a^3-b^3 = (a-b)(a^2+ab+b^2) \): \( x^3-1 = (x-1)(x^2+x+1) \).
Factor the denominator \( x^2+5x-6 \): \( (x-1)(x+6) \).
So, \( \lim_{x \to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+6)} \)
Since \( x \to 1 \), \( x \neq 1 \), so \( x-1 \neq 0 \). We can cancel \( (x-1) \) from numerator and denominator.
\( = \lim_{x \to 1} \frac{x^2+x+1}{x+6} \)
Substitute \( x=1 \):
\( = \frac{(1)^2+1+1}{1+6} \)
\( = \frac{1+1+1}{7} \)
\( = \frac{3}{7} \)
In simple words: We factored both the numerator and the denominator, identified and canceled the common factor \( (x-1) \) that caused the indeterminate form, and then substituted the limit value.

🎯 Exam Tip: For rational functions, if direct substitution leads to \( \frac{0}{0} \), factorize both the numerator and denominator to identify and cancel the common factor (often \( x-a \)). Remember common algebraic identities like \( a^3-b^3 \).

Question 5. Evaluate \( \lim_{x \to 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}} \)
Answer:
Solution:
We have \( \lim_{x \to 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}} \)
Multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{x-2}+\sqrt{4-x} \):
\( = \lim_{x \to 3} \frac{x-3}{\sqrt{x-2}-\sqrt{4-x}} \times \frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}} \)
Using \( (A-B)(A+B) = A^2-B^2 \) in the denominator:
\( = \lim_{x \to 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2})^2-(\sqrt{4-x})^2} \)
\( = \lim_{x \to 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)} \)
\( = \lim_{x \to 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{x-2-4+x} \)
\( = \lim_{x \to 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6} \)
Factor out 2 from the denominator: \( 2x-6 = 2(x-3) \).
\( = \lim_{x \to 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)} \)
Since \( x \to 3 \), \( x \neq 3 \), so \( x-3 \neq 0 \). We can cancel \( (x-3) \).
\( = \lim_{x \to 3} \frac{\sqrt{x-2}+\sqrt{4-x}}{2} \)
Substitute \( x=3 \):
\( = \frac{\sqrt{3-2}+\sqrt{4-3}}{2} \)
\( = \frac{\sqrt{1}+\sqrt{1}}{2} \)
\( = \frac{1+1}{2} \)
\( = \frac{2}{2} \)
\( = 1 \)
In simple words: We rationalized the denominator by multiplying by its conjugate, which helped eliminate the square roots and allowed us to cancel the problematic \( (x-3) \) term, leading to a direct substitution for the limit.

🎯 Exam Tip: When dealing with limits involving square roots in the denominator (or numerator) that lead to indeterminate forms, multiply by the conjugate to rationalize the expression. This technique often helps in simplifying the fraction.

Question 6. Evaluate \( \lim_{x \to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \)
Answer:
Solution:
We have \( \lim_{x \to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \)
Multiply by the conjugates of both the numerator and the denominator:
\( = \lim_{x \to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \times \frac{3+\sqrt{5+x}}{3+\sqrt{5+x}} \times \frac{1+\sqrt{5-x}}{1+\sqrt{5-x}} \)
\( = \lim_{x \to 4} \frac{(3^2 - (5+x))(1+\sqrt{5-x})}{(1^2 - (5-x))(3+\sqrt{5+x})} \)
\( = \lim_{x \to 4} \frac{(9 - 5 - x)(1+\sqrt{5-x})}{(1 - 5 + x)(3+\sqrt{5+x})} \)
\( = \lim_{x \to 4} \frac{(4 - x)(1+\sqrt{5-x})}{(-4 + x)(3+\sqrt{5+x})} \)
Notice that \( (4-x) = -(x-4) \) and \( (-4+x) = (x-4) \).
\( = \lim_{x \to 4} \frac{-(x-4)(1+\sqrt{5-x})}{(x-4)(3+\sqrt{5+x})} \)
Since \( x \to 4 \), \( x \neq 4 \), so \( x-4 \neq 0 \). We can cancel \( (x-4) \).
\( = \lim_{x \to 4} \frac{-(1+\sqrt{5-x})}{3+\sqrt{5+x}} \)
Substitute \( x=4 \):
\( = \frac{-(1+\sqrt{5-4})}{3+\sqrt{5+4}} \)
\( = \frac{-(1+\sqrt{1})}{3+\sqrt{9}} \)
\( = \frac{-(1+1)}{3+3} \)
\( = \frac{-2}{6} \)
\( = -\frac{1}{3} \)
In simple words: We rationalized both the numerator and the denominator by multiplying by their respective conjugates, simplified the resulting algebraic expressions, canceled the common factor, and then substituted the limit value.

🎯 Exam Tip: For limits involving roots in both the numerator and denominator, multiply by both conjugates. Be careful with algebraic simplification, especially with signs, before canceling terms. Always ensure you are simplifying the expression leading to the indeterminate form.

Question 7. Evaluate \( \lim_{x \to 0} \frac{5^x-1}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{5^x-1}{x} \)
This is a standard limit formula:
\( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \)
Here, \( a = 5 \).
Therefore, \( \lim_{x \to 0} \frac{5^x-1}{x} = \log 5 \)
In simple words: This limit is a direct application of a standard exponential limit formula, which states that the limit of \( (a^x-1)/x \) as \( x \) approaches 0 is \( \log a \).

🎯 Exam Tip: Memorize the standard limit formulas, especially for exponential and logarithmic functions. Recognizing the pattern can save significant time in calculations.

Question 8. Evaluate \( \lim_{x \to 0} \left(1+\frac{x}{5}\right)^{\frac{1}{x}} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \left(1+\frac{x}{5}\right)^{\frac{1}{x}} \)
To use the standard limit formula \( \lim_{y \to 0} (1+y)^{\frac{1}{y}} = e \), we need to manipulate the exponent.
Let \( y = \frac{x}{5} \). Then as \( x \to 0 \), \( y \to 0 \).
The exponent is \( \frac{1}{x} \). We want \( \frac{1}{y} = \frac{5}{x} \).
We can rewrite \( \frac{1}{x} \) as \( \frac{1}{5} \times \frac{5}{x} \).
So, \( \lim_{x \to 0} \left(1+\frac{x}{5}\right)^{\frac{1}{x}} = \lim_{x \to 0} \left(1+\frac{x}{5}\right)^{\frac{1}{5} \cdot \frac{5}{x}} \)
\( = \lim_{x \to 0} \left[\left(1+\frac{x}{5}\right)^{\frac{5}{x}}\right]^{\frac{1}{5}} \)
Since \( \lim_{x \to 0} \left(1+\frac{x}{5}\right)^{\frac{5}{x}} = e \),
\( = e^{\frac{1}{5}} \)
In simple words: We transformed the limit expression by adjusting the exponent to match the standard form \( (1+y)^{1/y} \) for the natural exponential 'e', then applied the limit.

🎯 Exam Tip: For limits of the form \( (1+f(x))^{1/f(x)} \) as \( f(x) \to 0 \), the limit is 'e'. Carefully adjust the exponent to create the exact reciprocal of the term added to 1, then apply the limit property for powers.

Question 9. Evaluate \( \lim_{x \to 0} \frac{\log(1+9x)}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{\log(1+9x)}{x} \)
To use the standard limit formula \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \), we need to modify the denominator.
Multiply and divide by 9 in the expression:
\( = \lim_{x \to 0} \frac{\log(1+9x)}{9x} \times 9 \)
Let \( y = 9x \). As \( x \to 0 \), \( y \to 0 \).
So, \( = 9 \lim_{y \to 0} \frac{\log(1+y)}{y} \)
\( = 9 \times 1 \)
\( = 9 \)
In simple words: We adjusted the denominator to match the argument inside the logarithm by multiplying and dividing by 9, then applied the standard logarithmic limit formula.

🎯 Exam Tip: Remember the standard logarithmic limit \( \lim_{x \to 0} \frac{\log(1+kx)}{x} = k \). This can be derived by multiplying and dividing by 'k' to align with the basic form \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \).

Question 10. Evaluate \( \lim_{x \to 0} \frac{(1-x)^5-1}{(1-x)^3-1} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{(1-x)^5-1}{(1-x)^3-1} \)
Let \( y = 1-x \). As \( x \to 0 \), \( y \to 1 \).
The limit becomes:
\( \lim_{y \to 1} \frac{y^5-1}{y^3-1} \)
We can divide both numerator and denominator by \( (y-1) \) to use the standard limit formula \( \lim_{y \to a} \frac{y^n - a^n}{y-a} = na^{n-1} \).
\( = \lim_{y \to 1} \frac{\frac{y^5-1}{y-1}}{\frac{y^3-1}{y-1}} \)
Applying the limit formula to the numerator and denominator separately:
Numerator: \( \lim_{y \to 1} \frac{y^5-1^5}{y-1} = 5(1)^{5-1} = 5 \)
Denominator: \( \lim_{y \to 1} \frac{y^3-1^3}{y-1} = 3(1)^{3-1} = 3 \)
Therefore, the limit is \( \frac{5}{3} \)
In simple words: We simplified the expression by substituting \( y = 1-x \), then applied the power rule for limits to both the numerator and the denominator after dividing them by \( (y-1) \).

🎯 Exam Tip: When faced with a fraction where both numerator and denominator lead to an indeterminate form and are of the \( x^n-a^n \) type, consider dividing both by \( (x-a) \) (or the equivalent transformed variable) to apply the limit formula separately.

Question 11. Evaluate \( \lim_{x \to 0} \frac{a^x+b^x+c^x-3}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{a^x+b^x+c^x-3}{x} \)
Rewrite \( -3 \) as \( (-1) + (-1) + (-1) \) and group terms:
\( = \lim_{x \to 0} \frac{(a^x-1)+(b^x-1)+(c^x-1)}{x} \)
Split the limit into three separate fractions:
\( = \lim_{x \to 0} \left(\frac{a^x-1}{x} + \frac{b^x-1}{x} + \frac{c^x-1}{x}\right) \)
Apply the limit to each term:
\( = \lim_{x \to 0} \frac{a^x-1}{x} + \lim_{x \to 0} \frac{b^x-1}{x} + \lim_{x \to 0} \frac{c^x-1}{x} \)
Using the standard limit formula \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \):
\( = \log a + \log b + \log c \)
Using the logarithm property \( \log A + \log B + \log C = \log(ABC) \):
\( = \log (abc) \)
In simple words: We rearranged the expression by subtracting 1 from each exponential term, then split the limit into individual standard exponential limit forms. Finally, we applied the logarithmic sum property to combine the results.

🎯 Exam Tip: For expressions like \( \frac{a^x+b^x+c^x-k}{x} \) where \( k \) is the number of terms, break down the constant \( k \) into \( 1+1+... \) and group it with each exponential term to use the standard limit \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \). Remember logarithm properties for simplification.

Question 12. Evaluate \( \lim_{x \to 0} \frac{e^x+e^{-x}-2}{x^2} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{e^x+e^{-x}-2}{x^2} \)
Rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\( = \lim_{x \to 0} \frac{e^x+\frac{1}{e^x}-2}{x^2} \)
Combine the terms in the numerator by finding a common denominator:
\( = \lim_{x \to 0} \frac{\frac{(e^x)^2+1-2e^x}{e^x}}{x^2} \)
The numerator is a perfect square, \( (e^x-1)^2 \):
\( = \lim_{x \to 0} \frac{(e^x-1)^2}{x^2 e^x} \)
Rearrange the terms to form a standard limit expression:
\( = \lim_{x \to 0} \left[\left(\frac{e^x-1}{x}\right)^2 \times \frac{1}{e^x}\right] \)
Apply the limit separately:
\( = \left(\lim_{x \to 0} \frac{e^x-1}{x}\right)^2 \times \left(\lim_{x \to 0} \frac{1}{e^x}\right) \)
Using the standard limit formula \( \lim_{x \to 0} \frac{e^x-1}{x} = \log e = 1 \):
And for the second part, substitute \( x=0 \): \( \frac{1}{e^0} = \frac{1}{1} = 1 \).
\( = (1)^2 \times 1 \)
\( = 1 \times 1 \)
\( = 1 \)
In simple words: We simplified the numerator by finding a common denominator and recognizing a perfect square, then rearranged the expression to utilize the standard limit formula for \( (e^x-1)/x \), and finally evaluated the remaining part by direct substitution.

🎯 Exam Tip: When \( e^x \) and \( e^{-x} \) appear together, consider combining them over a common denominator. Also, recognize common algebraic identities like \( (a-b)^2 \) to simplify complex expressions into standard limit forms.

Question 13. Evaluate \( \lim_{x \to 0} \frac{x(6^x-3^x)}{(2^x-1)\log(1+x)} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{x(6^x-3^x)}{(2^x-1)\log(1+x)} \)
Factor out \( 3^x \) from \( (6^x-3^x) \): \( 6^x-3^x = 3^x(2^x-1) \).
\( = \lim_{x \to 0} \frac{x \cdot 3^x(2^x-1)}{(2^x-1)\log(1+x)} \)
Since \( x \to 0 \), \( 2^x \to 2^0 = 1 \), so \( 2^x-1 \neq 0 \). We can cancel \( (2^x-1) \).
\( = \lim_{x \to 0} \frac{x \cdot 3^x}{\log(1+x)} \)
Rearrange the terms to align with standard limit forms:
\( = \lim_{x \to 0} \frac{3^x}{\frac{\log(1+x)}{x}} \)
Apply the limit to the numerator and denominator separately:
\( = \frac{\lim_{x \to 0} 3^x}{\lim_{x \to 0} \frac{\log(1+x)}{x}} \)
Using standard limit formulas: \( \lim_{x \to 0} 3^x = 3^0 = 1 \) and \( \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 \).
\( = \frac{1}{1} \)
\( = 1 \)
In simple words: We factored the exponential terms, canceled the common factor, then rearranged the expression to match standard limit forms for exponential and logarithmic functions, leading to a simple division of limits.

🎯 Exam Tip: Look for opportunities to factorize and simplify exponential terms (e.g., \( a^x - b^x = b^x((\frac{a}{b})^x - 1) \)). This often reveals common factors that can be canceled, reducing the expression to a combination of standard limits.

Question 14. Evaluate \( \lim_{x \to 0} \frac{a^{3x}-a^{2x}-a^x+1}{x^2} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{a^{3x}-a^{2x}-a^x+1}{x^2} \)
Factorize the numerator by grouping terms:
\( = \lim_{x \to 0} \frac{a^{2x}(a^x-1) - 1(a^x-1)}{x^2} \)
\( = \lim_{x \to 0} \frac{(a^x-1)(a^{2x}-1)}{x^2} \)
Separate the fraction into two parts to utilize standard limit formulas:
\( = \lim_{x \to 0} \left(\frac{a^x-1}{x} \times \frac{a^{2x}-1}{x}\right) \)
Apply the limit to each part:
\( = \left(\lim_{x \to 0} \frac{a^x-1}{x}\right) \times \left(\lim_{x \to 0} \frac{a^{2x}-1}{x}\right) \)
For the second term, we need to multiply and divide by 2:
\( = \left(\lim_{x \to 0} \frac{a^x-1}{x}\right) \times \left(\lim_{x \to 0} \frac{a^{2x}-1}{2x} \times 2\right) \)
Using the standard limit formula \( \lim_{y \to 0} \frac{a^y-1}{y} = \log a \):
\( = (\log a) \times (\log a \times 2) \)
\( = (\log a) \times (2 \log a) \)
\( = 2(\log a)^2 \)
In simple words: We factored the complex numerator into two simpler exponential terms, then split the fraction and applied the standard exponential limit formula to each part, remembering to adjust the second term to match the formula.

🎯 Exam Tip: When \( x^2 \) is in the denominator, look to factorize the numerator into two terms, each containing \( x \) in its denominator. Remember to adjust coefficients (like the 2 for \( a^{2x} \)) to match standard limit forms. Keep track of \( (\log a)^2 \) vs \( \log(a^2) \).

Question 15. Evaluate \( \lim_{x \to 0} \frac{(5^x-1)^2}{x \cdot \log(1+x)} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{(5^x-1)^2}{x \cdot \log(1+x)} \)
To use standard limit formulas, divide both the numerator and the denominator by \( x^2 \):
\( = \lim_{x \to 0} \frac{\frac{(5^x-1)^2}{x^2}}{\frac{x \cdot \log(1+x)}{x^2}} \)
\( = \lim_{x \to 0} \frac{\left(\frac{5^x-1}{x}\right)^2}{\frac{\log(1+x)}{x}} \)
Apply the limit to the numerator and denominator separately:
\( = \frac{\lim_{x \to 0} \left(\frac{5^x-1}{x}\right)^2}{\lim_{x \to 0} \frac{\log(1+x)}{x}} \)
Using the standard limit formulas: \( \lim_{x \to 0} \frac{a^x-1}{x} = \log a \) and \( \lim_{x \to 0} \frac{\log(1+x)}{x} = 1 \).
Here, \( a=5 \).
\( = \frac{(\log 5)^2}{1} \)
\( = (\log 5)^2 \)
In simple words: We transformed the expression by dividing both the numerator and denominator by \( x^2 \) to align with standard exponential and logarithmic limit forms, then applied these formulas to evaluate the limit.

🎯 Exam Tip: When you see \( x^n \) in the denominator and terms that simplify to standard limit forms after division by \( x \), consider dividing both the numerator and denominator by \( x^n \). This strategy helps break down complex limits into recognizable standard forms.

Question 16. Evaluate \( \lim_{x \to 0} \frac{a^{4x}-1}{b^{2x}-1} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{a^{4x}-1}{b^{2x}-1} \)
To use the standard limit formula \( \lim_{y \to 0} \frac{k^y-1}{y} = \log k \), divide both the numerator and the denominator by \( x \):
\( = \lim_{x \to 0} \frac{\frac{a^{4x}-1}{x}}{\frac{b^{2x}-1}{x}} \)
Now, adjust the denominators to match the exponent coefficients:
\( = \lim_{x \to 0} \frac{\frac{a^{4x}-1}{4x} \times 4}{\frac{b^{2x}-1}{2x} \times 2} \)
Apply the limit to the numerator and denominator separately:
\( = \frac{\lim_{x \to 0} \left(\frac{a^{4x}-1}{4x}\right) \times 4}{\lim_{x \to 0} \left(\frac{b^{2x}-1}{2x}\right) \times 2} \)
Using the standard limit formula \( \lim_{y \to 0} \frac{k^y-1}{y} = \log k \):
\( = \frac{(\log a) \times 4}{(\log b) \times 2} \)
\( = \frac{4 \log a}{2 \log b} \)
\( = \frac{2 \log a}{\log b} \)
In simple words: We manipulated the expression by dividing the numerator and denominator by \( x \), then adjusted each part to match the standard exponential limit form by multiplying and dividing by the exponent's coefficient, finally applying the limit.

🎯 Exam Tip: When the numerator and denominator are both of the form \( (k^x-1) \), divide both by \( x \) and then adjust the constants to fit the standard formula \( \lim_{y \to 0} \frac{a^y-1}{y} = \log a \). Ensure you multiply by the compensating factor for the adjustment.

Question 17. Evaluate \( \lim_{x \to 0} \frac{\log 100+\log(0.01+x)}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{\log 100+\log(0.01+x)}{x} \)
Use the logarithm property \( \log A + \log B = \log(AB) \):
\( = \lim_{x \to 0} \frac{\log[100 \times (0.01+x)]}{x} \)
Distribute 100 inside the parenthesis:
\( = \lim_{x \to 0} \frac{\log[1+100x]}{x} \)
To use the standard limit \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \), multiply and divide the denominator by 100:
\( = \lim_{x \to 0} \frac{\log(1+100x)}{100x} \times 100 \)
Let \( y = 100x \). As \( x \to 0 \), \( y \to 0 \).
\( = 100 \times \lim_{y \to 0} \frac{\log(1+y)}{y} \)
\( = 100 \times 1 \)
\( = 100 \)
In simple words: We used logarithm properties to combine the terms in the numerator, simplified the expression to a standard logarithmic limit form, and then applied the formula to find the value.

🎯 Exam Tip: Familiarize yourself with logarithm properties (\( \log A + \log B = \log(AB) \), \( \log A - \log B = \log(A/B) \)) as they are crucial for simplifying expressions before applying limit formulas. Remember the standard limit \( \lim_{x \to 0} \frac{\log(1+kx)}{x} = k \).

Question 18. Evaluate \( \lim_{x \to 0} \frac{\log(4-x)-\log(4+x)}{x} \)
Answer:
Solution:
We have \( \lim_{x \to 0} \frac{\log(4-x)-\log(4+x)}{x} \)
Use the logarithm property \( \log A - \log B = \log(A/B) \):
\( = \lim_{x \to 0} \frac{\log\left(\frac{4-x}{4+x}\right)}{x} \)
Alternatively, using \( \log(A) = \log(k(A/k)) = \log k + \log(A/k) \):
Numerator: \( \log(4-x) - \log(4+x) = [\log 4 + \log(1-\frac{x}{4})] - [\log 4 + \log(1+\frac{x}{4})] \)
\( = \log(1-\frac{x}{4}) - \log(1+\frac{x}{4}) \)
So, the limit becomes:
\( = \lim_{x \to 0} \frac{\log(1-\frac{x}{4}) - \log(1+\frac{x}{4})}{x} \)
Split the fraction into two terms:
\( = \lim_{x \to 0} \left(\frac{\log(1-\frac{x}{4})}{x} - \frac{\log(1+\frac{x}{4})}{x}\right) \)
Adjust denominators to match the arguments inside the logarithms:
\( = \lim_{x \to 0} \left(\frac{\log(1-\frac{x}{4})}{-\frac{x}{4}} \times (-\frac{1}{4})\right) - \lim_{x \to 0} \left(\frac{\log(1+\frac{x}{4})}{\frac{x}{4}} \times (\frac{1}{4})\right) \)
Using the standard limit formula \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \):
\( = (1) \times (-\frac{1}{4}) - (1) \times (\frac{1}{4}) \)
\( = -\frac{1}{4} - \frac{1}{4} \)
\( = -\frac{2}{4} \)
\( = -\frac{1}{2} \)
In simple words: We used logarithm properties to expand and simplify the numerator, then split the expression and adjusted each term to match the standard logarithmic limit form. Finally, we applied the limit formula and combined the results.

🎯 Exam Tip: For logarithmic differences like \( \log(A-x) - \log(A+x) \), expand using \( \log(k \pm x) = \log k + \log(1 \pm x/k) \). This often simplifies the expression by canceling \( \log k \) terms, making it easier to apply the standard limit formula \( \lim_{y \to 0} \frac{\log(1+y)}{y} = 1 \).

Question 19. Evaluate the limit of the function if it exists at \( x = 1 \) where,
\[ f(x) = \begin{cases} 7-4x & x < 1 \\ x^2+2 & x \ge 1 \end{cases} \]
Answer:
Solution:
For the limit to exist at \( x=1 \), the left-hand limit (LHL) and the right-hand limit (RHL) must be equal.

1. Left-Hand Limit (LHL):
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (7-4x) \)
Substitute \( x=1 \):
\( = 7-4(1) \)
\( = 7-4 \)
\( = 3 \)

2. Right-Hand Limit (RHL):
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2+2) \)
Substitute \( x=1 \):
\( = (1)^2+2 \)
\( = 1+2 \)
\( = 3 \)

Since LHL \( = \) RHL \( = 3 \), the limit of the function exists at \( x=1 \).
Therefore, \( \lim_{x \to 1} f(x) = 3 \)
In simple words: We evaluated the function's limit from both the left and the right sides of \( x=1 \). Since both approaches yielded the same value, the limit exists and is equal to that value.

🎯 Exam Tip: For piecewise functions, evaluating the limit at the boundary point requires checking both the left-hand limit (LHL) and the right-hand limit (RHL). The limit exists only if LHL = RHL. Always clearly state LHL and RHL calculations.