CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 04

TYPE: 7

Limits of Exponential and Logarithmic Functions

Q.1) Evaluate \(\lim_{x \to 0} \left( \frac{2^x - 1}{\sqrt{1+x} - 1} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{2^x - 1}{\sqrt{1+x} - 1} \right)\)
Rationalize the denominator:
\(= \lim_{x \to 0} \left( \frac{2^x - 1}{\sqrt{1+x} - 1} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} \right)\)
\(= \lim_{x \to 0} \left( \frac{(2^x - 1)(\sqrt{1+x} + 1)}{(1+x) - 1} \right)\)
\(= \lim_{x \to 0} \left( \frac{2^x - 1}{x} \right) \times \lim_{x \to 0} (\sqrt{1+x} + 1)\)
\(= \log 2 \times (1 + 1)\)
\(= 2 \log 2\) ans.
In simple words: Multiply the numerator and denominator by the conjugate of the denominator, then apply standard limit formulas for exponential functions.

Exam Tip: Use the standard limit result \(\lim_{x \to 0} \frac{a^x - 1}{x} = \log a\) to simplify exponential terms quickly.

Q.2) Evaluate \(\lim_{x \to 0} \left( \frac{10^x - 2^x - 5^x + 1}{x \tan x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{10^x - 2^x - 5^x + 1}{x \tan x} \right)\)
Since \(10^x = 2^x \cdot 5^x\), we can factor the numerator:
\(= \lim_{x \to 0} \left( \frac{2^x(5^x - 1) - 1(5^x - 1)}{x \tan x} \right)\)
\(= \lim_{x \to 0} \left( \frac{(5^x - 1)(2^x - 1)}{x \tan x} \right)\)
Divide numerator and denominator by \(x^2\):
\(= \lim_{x \to 0} \left( \frac{\left(\frac{5^x - 1}{x}\right) \left(\frac{2^x - 1}{x}\right)}{\frac{\tan x}{x}} \right)\)
\(= \frac{\lim_{x \to 0} \left(\frac{5^x - 1}{x}\right) \times \lim_{x \to 0} \left(\frac{2^x - 1}{x}\right)}{\lim_{x \to 0} \left(\frac{\tan x}{x}\right)}\)
\(= \frac{(\log 5)(\log 2)}{1}\)
\(= \log 5 \log 2\) ans.
In simple words: Factor by grouping to split the expression into standard limit forms, then divide by \(x^2\) to apply standard formulas.

Exam Tip: Whenever you see mixed bases like \(10^x\), express them as products of prime bases like \(2^x\) and \(5^x\) to enable factoring.

Q.3) Evaluate \(\lim_{x \to 0} \left( \frac{e^x + e^{-x} - 2}{x^2} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{e^x + e^{-x} - 2}{x^2} \right)\)
Rewrite \(e^{-x}\) as \(\frac{1}{e^x}\):
\(= \lim_{x \to 0} \left( \frac{e^x + \frac{1}{e^x} - 2}{x^2} \right)\)
Take the LCM inside the numerator:
\(= \lim_{x \to 0} \left( \frac{e^{2x} + 1 - 2e^x}{e^x \cdot x^2} \right)\)
Recognize the numerator as a complete square: \((e^x - 1)^2\)
\(= \lim_{x \to 0} \left( \frac{(e^x - 1)^2}{e^x \cdot x^2} \right)\)
\(= \lim_{x \to 0} \left( \left(\frac{e^x - 1}{x}\right)^2 \times \frac{1}{e^x} \right)\)
\(= (1)^2 \times \frac{1}{e^0}\)
\(= 1\) ans.
In simple words: Convert negative exponents to fractions, take a common denominator, and factor the resulting quadratic form into a squared term.

Exam Tip: The combination \(e^x + e^{-x} - 2\) can always be rewritten as \((e^x - 1)^2 / e^x\) for easy evaluation.

Q.4) Evaluate \(\lim_{x \to 0} \left( \frac{\log(1 + x^3)}{\sin^3 x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{\log(1 + x^3)}{\sin^3 x} \right)\)
Multiply and divide by \(x^3\):
\(= \lim_{x \to 0} \left( \frac{\frac{\log(1 + x^3)}{x^3} \times x^3}{\frac{\sin^3 x}{x^3} \times x^3} \right)\)
\(= \frac{\lim_{x \to 0} \left(\frac{\log(1 + x^3)}{x^3}\right)}{\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^3}\)
\(= \frac{1}{1^3} = 1\) ans.
In simple words: Divide both numerator and denominator by \(x^3\) to match the standard logarithmic and trigonometric limits.

Exam Tip: Remember that \(\lim_{x \to 0} \frac{\log(1 + u)}{u} = 1\) when \(u \to 0\).

Q.5) Evaluate \(\lim_{x \to 0} \left( \frac{3^{2x} - 2^{3x}}{\sin(3x)} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{3^{2x} - 2^{3x}}{\sin(3x)} \right)\)
Note that \(3^{2x} = 9^x\) and \(2^{3x} = 8^x\). Add and subtract 1 in the numerator:
\(= \lim_{x \to 0} \left( \frac{9^x - 8^x - 1 + 1}{\sin(3x)} \right)\)
\(= \lim_{x \to 0} \left( \frac{(9^x - 1) - (8^x - 1)}{\sin(3x)} \right)\)
Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \left( \frac{\left(\frac{9^x - 1}{x}\right) - \left(\frac{8^x - 1}{x}\right)}{\frac{\sin(3x)}{3x} \times 3} \right)\)
\(= \frac{\log 9 - \log 8}{3}\)
\(= \frac{\log(9/8)}{3} = \frac{1}{3} \log\left(\frac{9}{8}\right)\) ans.
In simple words: Adjust powers, introduce \(-1\) and \(+1\) to create standard exponential forms, and divide by \(x\).

Exam Tip: Use logarithm properties like \(\log A - \log B = \log(\frac{A}{B})\) to combine log terms in the final step.

Q.6) Evaluate \(\lim_{x \to 0} \left( \frac{x(e^x - 1)}{1 - \cos x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{x(e^x - 1)}{1 - \cos x} \right)\)
Use the trigonometric identity \(1 - \cos x = 2 \sin^2(x/2)\):
\(= \lim_{x \to 0} \left( \frac{x(e^x - 1)}{2 \sin^2(x/2)} \right)\)
Multiply and divide the denominator by \(x^2 / 4\):
\(= \lim_{x \to 0} \left( \frac{\frac{e^x - 1}{x}}{\frac{2 \sin^2(x/2)}{(x/2)^2} \times \frac{1}{4}} \right)\)
\(= \frac{2 \lim_{x \to 0} \left(\frac{e^x - 1}{x}\right)}{\lim_{x \to 0} \left(\frac{\sin(x/2)}{x/2}\right)^2}\)
\(= \frac{2(1)}{1} = 2\) ans.
In simple words: Replace \(1 - \cos x\) with its half-angle formula and divide appropriately by powers of \(x\) to apply standard limits.

Exam Tip: The standard limit for cosine denominator is \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\).

Q.7) Evaluate \(\lim_{x \to 0} \left( \frac{a^x + b^x + c^x - 3}{x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{a^x + b^x + c^x - 3}{x} \right)\)
Split the number 3 into three 1s:
\(= \lim_{x \to 0} \left( \frac{a^x + b^x + c^x - 1 - 1 - 1}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{(a^x - 1) + (b^x - 1) + (c^x - 1)}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) + \lim_{x \to 0} \left( \frac{b^x - 1}{x} \right) + \lim_{x \to 0} \left( \frac{c^x - 1}{x} \right)\)
\(= \log a + \log b + \log c = \log(abc)\) ans.
In simple words: Group each exponential term with 1, split the fraction into three parts, and apply the standard exponential limit formula to each.

Exam Tip: Sum of logarithms equals the logarithm of the product: \(\log a + \log b + \log c = \log(abc)\).

Q.8) Evaluate \(\lim_{x \to 0} \left( \frac{a^x + b^x - c^x - d^x}{x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{a^x + b^x - c^x - d^x}{x} \right)\)
Add and subtract appropriate 1s:
\(= \lim_{x \to 0} \left( \frac{(a^x - 1) + (b^x - 1) - (c^x - 1) - (d^x - 1)}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) + \lim_{x \to 0} \left( \frac{b^x - 1}{x} \right) - \lim_{x \to 0} \left( \frac{c^x - 1}{x} \right) - \lim_{x \to 0} \left( \frac{d^x - 1}{x} \right)\)
\(= \log a + \log b - \log c - \log d\)
\(= (\log a + \log b) - (\log c + \log d)\)
\(= \log(ab) - \log(cd) = \log\left(\frac{ab}{cd}\right)\) ans.
In simple words: Pair each exponential term with 1, separate into individual limits, and combine using logarithm laws.

Exam Tip: Positive log terms form the numerator, and negative log terms form the denominator inside a single logarithm.

Q.9) Evaluate \(\lim_{x \to 0} \left( \frac{\log(5 + x) - \log(5 - x)}{x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{\log(5 + x) - \log(5 - x)}{x} \right)\)
Factor out 5 from each argument inside the logarithms:
\(= \lim_{x \to 0} \left( \frac{\log\left(5(1 + \frac{x}{5})\right) - \log\left(5(1 - \frac{x}{5})\right)}{x} \right)\)
Using log properties \(\log(xy) = \log x + \log y\):
\(= \lim_{x \to 0} \left( \frac{(\log 5 + \log(1 + \frac{x}{5})) - (\log 5 + \log(1 - \frac{x}{5}))}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{\log(1 + \frac{x}{5}) - \log(1 - \frac{x}{5})}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{\log(1 + \frac{x}{5})}{\frac{x}{5} \times 5} \right) + \lim_{x \to 0} \left( \frac{\log(1 - \frac{x}{5})}{-\frac{x}{5} \times -5} \right)\)
\(= \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\) ans.
In simple words: Factor out 5 from the log arguments, cancel \(\log 5\), and adjust the denominators to apply \(\lim_{x \to 0} \frac{\log(1+u)}{u} = 1\).

Exam Tip: Always factor out the constant term inside logarithmic expressions before applying standard limits.

Q.10) Evaluate \(\lim_{x \to 0} \left( \frac{e^{3+x} - \sin x - e^3}{x} \right)\)
Answer: We have \(\lim_{x \to 0} \left( \frac{e^{3+x} - \sin x - e^3}{x} \right)\)
Rewrite \(e^{3+x}\) as \(e^3 \cdot e^x\):
\(= \lim_{x \to 0} \left( \frac{e^3 e^x - e^3 - \sin x}{x} \right)\)
Factor out \(e^3\) from the first two terms:
\(= \lim_{x \to 0} \left( \frac{e^3(e^x - 1) - \sin x}{x} \right)\)
Split into separate fractions:
\(= \lim_{x \to 0} \left( e^3 \left(\frac{e^x - 1}{x}\right) - \frac{\sin x}{x} \right)\)
\(= e^3 \lim_{x \to 0} \left(\frac{e^x - 1}{x}\right) - \lim_{x \to 0} \left(\frac{\sin x}{x}\right)\)
\(= e^3(1) - (1) = e^3 - 1\) ans.
In simple words: Separate terms containing \(e^3\), factor out the constant, and apply standard limits for exponentials and trigonometry.

Exam Tip: Index laws like \(a^{m+n} = a^m \cdot a^n\) help separate exponential terms and reveal common factors.