First slide

Introduction to limits

Question

$Let f : R \to R be such that f (1) = 3 and f^{'} (1) = 6, then \lim_{x \to 0} {(\frac{f (1 + x)}{f (1)})}^{1 / x} is equal to$

Moderate

A

1
B

$e^{\frac{1}{2}}$
C

$e^{2}$
D

$e^{3}$

Solution

$Given that f : R \to R such that$
$\underset{}{f} (1) = 3 and f^{'} (1) = 6$
$Then \lim_{x \to 0} {[\frac{f (1 + x)}{f (1)}]}^{1 / x} = e^{\lim_{x \to 0} \frac{1}{x} [logf (1 + x) - \log f (1)]}$
$= e^{\lim_{x \to 0} \frac{1}{f (1 + x)} f^{1} (1 + x)} 1 [Using L^{'} Hospital rule]$
$= e^{2}$

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