Exponential and Logarithmic Limits: Formula and Examples

Exponential Limits - Definition, Domain, Range and Standard 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 $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = a^x$, where $a > 0$ and $a \neq 1$, is called an exponential function.

The domain of an exponential function is $(-\infty, \infty)$ and the range is $(0, \infty)$ since $a^x$ always takes positive values.

Behaviour of $f(x) = a^x$ Based on the Base $a$

Since $a > 0$ and $a \neq 1$, two cases arise:

Case I: When $a > 1$ (Exponential Growth)

As $x$ increases, $a^x$ also increases:

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

Case II: When $0 < a < 1$ (Exponential Decay)

As $x$ increases, $a^x$ decreases, but $f(x) > 0$ for all $x \in \mathbb{R}$.

Standard Limits Involving Exponential Functions

Limit of $a^x$ as $x \to 0$:

$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$

Proof using Taylor series expansion of $a^x$:

$\lim_{x \to 0} \frac{a^x - 1}{x} = \lim_{x \to 0} \frac{1 + \frac{x (\ln a)}{1!} + \frac{x^2 (\ln a)^2}{2!} + \cdots - 1}{x}$

$= \lim_{x \to 0} \left( \frac{\ln a}{1!} + \frac{x (\ln a)^2}{2!} + \cdots \right) = \ln a$

Special case for $e^x$:

$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

General Result for Functions Approaching Zero

If $\lim_{x \to a} f(x) = 0$, then:

(a) $\lim_{x \to a} \frac{a^{f(x)} - 1}{f(x)} = \ln a$

(b) $\lim_{x \to a} \frac{e^{f(x)} - 1}{f(x)} = \ln e = 1$

Logarithmic Limits

To evaluate logarithmic limits, we use the following standard result:

$\lim_{x \to 0} \frac{\log_e(1+x)}{x} = 1$

Proof using Taylor series expansion of $\log_e(1+x)$:

$\lim_{x \to 0} \frac{\log_e(1+x)}{x} = \lim_{x \to 0} \frac{x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots}{x}$

$= \lim_{x \to 0} \left( 1 - \frac{x}{2} + \frac{x^2}{3} - \cdots \right) = 1$

General Result for Logarithmic Limits

If $\lim_{x \to a} f(x) = 0$, then:

$\lim_{x \to a} \frac{\log_e(1 + f(x))}{f(x)} = 1$

Important formulae related to Exponential and Logarithmic Limits

Here we summarize all key formulas and standard results for exponential and logarithmic limits, including $\lim_{x \to 0} \frac{a^x - 1}{x}$, $\lim_{x \to 0} \frac{\log_e(1+x)}{x}$, and general forms, serving as a quick reference for students.

Solved Examples Based On Exponential Limits and Logarithmic 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.

List of topics related to Limits

We have provided below the list of topics related to limits, so that you can learn them effectively and use them in problems in the exam.

Indeterminate forms of limits

Algebra of Limits

Limits using expansion

Left and Right Hand Limits

Higher Order Derivatives

NCERT Resources

Here, we provide a collection of NCERT notes, solutions, and exemplar problems specifically for Limits and Derivatives, making it easier for students to follow the official syllabus and strengthen core concepts.

NCERT Notes for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Exempar Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

Practice Questions based on Exponential and Logarithmic Limits

This section includes practice questions questions on exponential and logarithmic limits, designed to enhance problem-solving speed, accuracy, and exam readiness for Class 11, Class 12, and competitive exams.

Exponential And Logarithmic Limits - Practice Question MCQ

We have shared the links below to practice questions on the related topics to limits: