JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 15)
$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$ is equal to
$$\frac{-2}{e}$$
$$e-e^2$$
0
$$e$$
Explanation
$$\lim _\limits{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$$
Using expansion
$$\begin{aligned} & =\lim _\limits{x \rightarrow 0} \frac{e-e\left[1-\frac{2 x}{2}+\frac{11 \times 4 x^2}{24}+\ldots\right]}{x} \\ & =\lim _\limits{x \rightarrow 0}\left(e-\frac{11 x}{6} e+\ldots\right)=e \end{aligned}$$
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