Q41 / 155 · WBJEE 2019 · Mathematics

$\lim _{x \rightarrow 0^{+}}\left(x^{n} \ln x\right), n>0$

  • A
    does not exist
  • Correct answer
    B
    exists and is zero
  • C
    exists and is 1
  • D
    exists and is $e^{-1}$

Solution

We have, $\lim _{x \rightarrow 0^{+}} x^{n} \ln x$ $=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{x^{-n}}$ $=\lim _{x \rightarrow 0^{+}} \frac{x}{-n x^{-n-1}} \quad$ [using L'Hospital's rule $]$ $=\lim _{x \rightarrow 0^{+}} \frac{-1}{n x^{-n}}=0 \quad\left[\because \frac{1}{x-n}=0,\right.$ when $\left.x=0\right]$

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

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