Q88 / 155 · WBJEE 2019 · Mathematics

$\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

  • A
    Does not exist finitely
  • B
    is 1
  • Correct answer
    C
    is $e^{2}$
  • D
    is 2

Solution

$$ \begin{aligned} &\text { Let } L=\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}\\ &\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\log \left(e^{x}+x\right)}{x}\\ &\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{\left(e^{x}+x\right)} \cdot e^{x}+1}{1}\\ &\text { [using L' Hospital rule] }\\ &\Rightarrow \log L=\lim _{x \rightarrow 0^{+}} \frac{e^{x}+1}{e^{x}+x}\\ &\Rightarrow \log L=2\\ &\Rightarrow \quad L=e^{2} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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