Exponential and Logarithmic Limits - Practice Questions & MCQ

Exponential Limits

To solve the limit of the function involving the exponential function, we use the following standard results.

(i) $\lim _{\mathbf{x} \rightarrow \mathbf{0}} \frac{\mathbf{a}^{\mathbf{x}}-\mathbf{1}}{\mathbf{x}}=\log _{\mathrm{e}} \mathrm{a}$

Proof:

$
\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x(\log a)}{1!}+\frac{x^2(\log a)^2}{2!}+\cdots\right)-1}{x}
$

[using Taylor series expansion of $a^x$ ]

$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(\frac{\log a}{1!}+\frac{x(\log a)^2}{2!}+\cdots\right) \\
& =\log _e a
\end{aligned}
$

(ii) $\lim _{\mathrm{x} \rightarrow 0} \frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}}=1$

In General, if $\lim _{x \rightarrow a} f(x)=0$, then we have
(a) $\lim _{x \rightarrow a} \frac{a^{f(x)}-1}{f(x)}=\log _e a$
(b) $\lim _{x \rightarrow a} \frac{e^{f(x)}-1}{f(x)}=\log _e e=1$

Logarithmic  Limits

To evaluate the Logarithmic limit we use the following results:

$
\lim _{\mathbf{x} \rightarrow \mathbf{0}} \frac{\log _e(\mathbf{1 + x})}{\mathbf{x}}=\mathbf{1}
$
Proof:

$
\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=\lim _{x \rightarrow 0} \frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots}{x}
$

[using Taylor series expansion of $\log _e(1+x)$ ]

$
\begin{aligned}
& =\lim _{x \rightarrow 0}\left(1-\frac{x}{2}+\frac{x^2}{3}-\cdots\right) \\
& =1
\end{aligned}
$
In General, if $\lim _{x \rightarrow a} f(x)=0$, then we have $\lim _{x \rightarrow a} \frac{\log _e(1+f(x))}{f(x)}=1$