First slide

Introduction to limits

Question

$lim_{x \to 1} \frac{\sin (e^{x - 1} - 1)}{\log x}$ is equal to

Moderate

A

1
B

0
C

e
D

$e^{- 1}$

Solution

We have,
$\begin{array}{l} lim_{x \to 1} \frac{\sin (e^{x - 1} - 1)}{\log x} = lim_{h \to 0} \frac{\sin (e^{h} - 1)}{\log (1 + h)} \\ = lim_{h \to 0} \frac{\sin (e^{h} - 1)}{(e^{h} - 1)} \times \frac{(e^{h} - 1)}{h} \times \frac{h}{\log (1 + h)} = 1 . \end{array}$

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