Medium MCQ +4 / -1 PYQ · JEE Mains 2024

$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to

  1. A $\frac{-2}{e}$
  2. B $e-e^2$
  3. C 0
  4. D $e$ Correct answer

Solution

<p>$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$</p> <p>Using expansion</p> <p>$$\begin{aligned} & =\lim _\limits{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\ & =\lim _\limits{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e \end{aligned}$$</p>

About this question

Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results

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