Question
Evaluate $\lim _{x \rightarrow 0} \frac{e^{k x}-1}{x}$.

Answer

We have, $\lim _{x \rightarrow 0} \frac{e^{b x}-1}{x}=\lim _{x \rightarrow 0} \frac{e^{b_x}-1}{x} \times \frac{b}{b}$
[multiplying numerator and denominator by b]
$=\lim _{x \rightarrow 0} \frac{b\left(e^{b x}-1\right)}{b x}$
On putting $h = bx$ and as $x \rightarrow 0$, then $h \rightarrow 0$, we get
$\begin{array}{l}\lim _{x \rightarrow 0} \frac{e^x-1}{x}=b \lim _{h \rightarrow 0} \frac{\left(e^h-1\right)}{h} \\ =b \times 1\left[\because \lim _{x \rightarrow 0} \frac{e^x-1}{x}=1\right] \\ = b \end{array}$

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