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) $\underset{\mathbf{x}\to \mathbf{0}}{lim}\frac{{\mathbf{a}}^{\mathbf{x}}-\mathbf{1}}{\mathbf{x}}={\log }_{\mathrm{e}}\mathrm{a}$

Proof:

$\underset{x\to 0}{lim}\frac{a^x-1}{x}=\underset{x\to 0}{lim}\frac{(1+\frac{x(\log a)}{1!}+\frac{x^2(\log a{)}^2}{2!}+\cdots )-1}{x}$

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

$\begin{array}{rl}& =\underset{x\to 0}{lim}(\frac{\log a}{1!}+\frac{x(\log a{)}^2}{2!}+\cdots )\\ & ={\log }_{e}a\end{array}$

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

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

Logarithmic  Limits

To evaluate the Logarithmic limit we use the following results:

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

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

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

$\begin{array}{rl}& =\underset{x\to 0}{lim}(1-\frac{x}{2}+\frac{x^2}{3}-\cdots )\\ & =1\end{array}$
In General, if $\underset{x\to a}{lim}f(x)=0$, then we have $\underset{x\to a}{lim}\frac{{\log }_e(1+f(x))}{f(x)}=1$