Q76 / 200 · Manipal (MET) 2012 · Mathematics

Let $f: R \rightarrow R$ be such that $f(1)=3$ and $f^{\prime}(1)=6$. Then, $\lim _{x \rightarrow 0}\left[\frac{f(1+x)}{f(1)}\right]^{1 / x}$ equals

  • A
    1
  • B
    $e^{1 / 2}$
  • Correct answer
    C
    $e^2$
  • D
    $e^3$

Solution

$\begin{aligned} & \text { Let } \quad y=\left[\frac{f(1+x)}{f(1)}\right]^{1 / x} \\ & \Rightarrow \quad \log y=\frac{1}{x}[\log f(1+x)-\log f(1)] \\ & \Rightarrow \lim _{x \rightarrow 0} \log y=\lim _{x \rightarrow 0} \frac{[\log f(1+x)-\log f(1)]}{x} \\ & \Rightarrow \lim _{x \rightarrow 0} \log y=\lim _{x \rightarrow 0}\left[\frac{1}{f(1+x)} f^{\prime}(1+x)\right] \\ & \text { (using L' Hospital's rule) } \\ & =\frac{f^{\prime}(1)}{f(1)}=\frac{6}{3}=2 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} y=e^2 \end{aligned}$

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

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