Exponential and Logarithmic Limits: Formula and Examples

Limits are one of the most basic ideas in calculus, where one can learn how functions behave as they approach particular points. Of interest, though, is that some limits tend not to be as straightforward as finding the others, such that they could evaluate the functions.The application of the Exponential Limits involves the behavior of exponential functions, such as $e^x$, as they approach infinity or other critical points.

In this article, we will cover the concept of the Exponential and Logarithmic Limits. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of five questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2015, one in 2020, and three in 2021.

Exponential Limits

The function which associates the number $a^x$ to each real number $x$ is called the exponential function and is denoted by $f(x)=a^x$

In other words, a function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=a^x$, where $\mathrm{a}>0$ and $a \neq 1$ is called the exponential function.

The domain of an exponential function is the set of all real numbers and the range of an exponential function is $(0, \infty)$ since it attains only positive values.

Now, as a $>0$ and $a \neq 1$, therefore the following two cases arise :
Case I: When $a>1$ :
Here we observe that as the values of $x$ increase, the values of $a^x$ also increase:

$f(x)$= $\begin{cases}<1 & \text { for } x<0 \\ 1 & \text { for } x=0 \\ >1 & \text { for } x>0\end{cases}$

Case II: When $0<a<1$ :

Here we observe that as the values of $x$ increase, the values of $a^x$ decrease and $f(x)>0$ for all $x$ $\varepsilon R$

To solve the limit of the function involving the exponential function, we use the following standard results:
(i) $\lim\limits _{\mathrm{x} \rightarrow 0} \frac{\mathrm{a}^{\mathrm{x}}-1}{\mathrm{x}}=\log _{\mathrm{e}} \mathrm{a}$

Proof:

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

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

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

Logarithmic Limits

To evaluate the Logarithmic limit we use the following results:

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

Proof:

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

Solved Examples Based On Exponential Limits

Example 1: Let $f: R \rightarrow R \space {\text {satisfy the equation }}$ $f(x+y)=f(x) \cdot f(y)_{\text {of all }} x, y \in R_{\text {and }} f(x) \neq 0$ for any $x \in R$. If the function f is differentiable at $\mathrm{x}=0$ and $\mathrm{f}^{\mathrm{f}}(0)=3$, then $\lim\limits _{h \rightarrow 0} \frac{1}{h}(f(h)-1)$ is equal to $\qquad$ [JEE Main 2021]

1) $3$
2) $5$
3) $2$
4) None of these

Solution:

$
\begin{aligned}
& f(x+y)=f(x) \cdot f(y) \text { then } \\
& \Rightarrow \quad f(x)=a^x
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow a=e^3 \\
& \therefore f(x)=\left(e^3\right)^x=e^{3 x}
\end{aligned}
$

Now, $\lim\limits _{h \rightarrow 0} \frac{f(h)-1}{h}=\lim\limits _{h \rightarrow 0}\left(\frac{e^{3 h}-1}{3 h} \times 3\right)=1 \times 3=3$

Hence, the answer is the $3 $

Example 2: If $\lim\limits _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{a x\left(e^{4 x}-1\right)}$ exists and is equal to b, then the value of $a-2 b$ is $\qquad$ - [JEE Main 2021]
1) $5$
2) $4$
3) $3$
4) $2$

Solution:

$\begin{equation}
\begin{aligned}
&\lim\limits _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{\operatorname{ax}\left(e^{4 x}-1\right)}\\
&\text { this is in } \frac{0}{0} \text { form }\\
&\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{\left(a x-\left(e^{4 x}-1\right)\right) 4 x}{\operatorname{ax}\left(e^{4 x}-1\right) 4 x} \\
& =\lim\limits _{x \rightarrow 0} \frac{\mathrm{ax}-\left(e^{4 x}-1\right)}{a x \cdot 4 x} \quad\left(\text { Use } \lim\limits _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1\right)
\end{aligned}
\end{aligned}
\end{equation}$

Apply L'Hopital rule

$=\lim\limits _{x \rightarrow 0} \frac{a-4 e^{4 x}}{8 a x} \quad\left(\frac{a-4}{0} \text { form }\right)
$

for the limit to exist, a = 4

$
\begin{aligned}
& =\lim\limits _{x \rightarrow 0} \frac{4-4 e^{4 x}}{32 x} \\
& =\lim\limits _{x \rightarrow 0} \frac{1-\mathrm{e}^{4 \mathrm{x}}}{8 \mathrm{x}} \quad\left(\frac{0}{0}\right) \\
& =\lim\limits _{x \rightarrow 0} \frac{-\mathrm{e}^{4 x} \cdot 4}{8}=-\frac{1}{2} \\
& \Rightarrow \mathrm{b}=-\frac{1}{2} \\
& a-2 b=4-2\left(-\frac{1}{2}\right) \\
& =5
\end{aligned}
$

Hence, the answer is the option 1.

Example 3: Find the value of $\lim\limits _{x \rightarrow 0} \frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1 / x\right)}-1\right)}{\sqrt{1+x^2+x^4}-1}$ [JEE Main 2020]
1) is equal to $1$
2) is equal to $0$
3) doesn't exist
4) is equal to $\sqrt{e}$

Solution:

$\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{\left(e^{\left(\sqrt{1+x^2+x^4}-1\right) / x}-1\right)}{\frac{\sqrt{1+x^2+x^4}-1}{x}} \\
& \lim\limits _{x \rightarrow 0} \frac{\sqrt{1+x^2+x^4}-1}{x} \quad\left(\frac{0}{0} \text { form }\right)
\end{aligned}
$

On applying L-Hopital rule, we get

$
\lim\limits _{x \rightarrow 0}\left(\frac{1\left(2 x+4 x^3\right)}{2 x \sqrt{1+x^2+x^4}}\right)=0
$

The limit is of the form $\lim\limits _{h \rightarrow 0} \frac{e^h-1}{h}$ It is equal to 1 .

Hence, the answer is the option 1 .

Example 4: If $\alpha, \beta$ are the distinct roots of $x^2+b x+c=0$, then $\lim\limits _{x \rightarrow \beta} \frac{e^{2\left(x^2+b x+c\right)}-1-2\left(x^2+b x+c\right)}{(x-\beta)^2}$ is equal to :
[JEE
Main 2021]

$
\begin{aligned}
& \text { 1) } 2\left(b^2+4 c\right) \\
& \text { 2) } b^2-4 c
\end{aligned}
$

3) $2\left(b^2-4 c\right)$

$
\text { 4) } b^2+4 c
$
Solution: $x^2+b x+c=(x-\alpha)(x-\beta)$

$
\begin{aligned}
& \lim\limits _{x \rightarrow \beta} \frac{e^{2(x-\alpha)(x-\beta)}-1-2(x-\alpha)(x-\beta)}{(x-\beta)^2} \text { Expanding } e^{2(x-\alpha)(x-\beta)} \\
& =\lim\limits _{x \rightarrow \beta} \frac{1+2(x-\alpha)(x-\beta)+\frac{(2(x-\alpha)(x-\beta))^2}{2!}+\cdots-1-2(x-\alpha)(x-\beta)}{(x-\beta)^2} \\
& =2(\alpha-\beta)^2 \\
& =2\left[(\alpha+\beta)^2-4 \alpha \beta\right]=2\left(b^2-4 c\right)
\end{aligned}
$

Hence, the answer is the option 3.

Example 5: Let k be a non - zero real number. If $ f(x)= \begin{cases}\frac{\left(e^x-1\right)^2}{\sin \left(\frac{x}{k}\right) \log \left(1+\frac{x}{4}\right)} & x \neq 0 \\ 12 & x=0 \text { is a continuous function, then }\end{cases}
$

the value of $k$ is :
[JEE Main 2015]

(1) $3$
2) $2$
3) $1$
4) $4$

Solution:

As we learned in
Evaluation of Exponential Limits -

$
\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{e^x-1}{x}=1 \\
& \lim\limits _{x \rightarrow 0} \frac{a^x-1}{x}=\log _e a \\
& \lim\limits _{x \rightarrow 0} \frac{e^{K x}-1}{x} \neq 1 \\
& \therefore \quad \lim\limits _{x \rightarrow 0} \frac{e^{K x}-1}{x} \times \frac{K}{K} \\
& \therefore \quad K \lim\limits _{x \rightarrow 0} \frac{e^{K x}-1}{K x}=K \times 1=K
\end{aligned}
$

- wherein

$
\lim\limits _{x \rightarrow 0} \frac{e^x-1}{x}
$

$x$ must be same

$
\begin{aligned}
& f(x)= \begin{cases}\frac{\left(e^x-1\right)^2}{\sin \left(\frac{x}{k}\right) \log \left(1+\frac{x}{4}\right)} & x \neq 0 \\
12 & x=0\end{cases} \\
& \therefore \lim\limits _{x \rightarrow \frac{0^{+}}{0^{-}}} \frac{\left(e^x-1\right)^2}{\sin \left(\frac{x}{k}\right) \log \left(1+\frac{x}{4}\right)} \\
& \therefore \lim\limits _{x \rightarrow \frac{0^{+}}{0^{-}}} \frac{\frac{\left(e^x-1\right)^2}{x^2}}{\frac{\sin \left(\frac{n}{k}\right) \log \left(1+\frac{r}{4}\right)}{\frac{x}{k} \times k} \frac{x}{4} \times 4} \\
& \lim\limits _{x \rightarrow 0} \frac{\left(\frac{e^x-1}{x}\right)^2}{\left(\frac{\sin \frac{x}{k}}{\frac{x}{k}}\right) \frac{\log \left(1+\frac{x}{4}\right)}{\frac{x}{4}}} \times 4 k \\
& \therefore \frac{1 \times 4 k}{1 \times 1}=4 k \\
& \therefore 4 k=12 \\
& \therefore k=3
\end{aligned}
$

Hence, the answer is the option 1.

Summary

Exponential limits and logarithms play a crucial role in understanding the behavior of exponential functions, such as $a^x$, as they approach infinity or other critical points. Calculus was created to describe how the quantities change. The concept of limit is the cornerstone on which the development of calculus rests.