Q99 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

$\lim _{x \rightarrow 0}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]=$

  • A
    0
  • B
    1
  • C
    2
  • Correct answer
    D
    $\frac{1}{2}$

Solution

$\lim _{x \rightarrow 0}\left[\frac{1}{x}-\frac{1}{e^x-1}\right]=\lim _{x \rightarrow 0}\left[\frac{e^x-1-x}{x e^x-x}\right]\left[\frac{0}{0} \text { form }\right]$
Applying L's Hopital rule $=\lim _{x \rightarrow 0}\left[\frac{e^x-1}{x e^x+e^x-1}\right] \quad\left[\frac{0}{0} \text { form }\right]$
Applying L's Hopital rule $=\lim _{x \rightarrow 0}\left[\frac{e^x}{x e^x+2 e^x}\right]=\frac{1}{2}$

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

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