Question

Find the value of given limit:-$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} =$
(a) $\dfrac{1}{2}$
(b) 0
(c) 1
(d) None

Answer 1

Here, we need to find the value of the given expression. We will simplify the value of the expression using rules of limits. Then, we will use substitution and L’Hopital’s rule to find the required value of the given expression.

Formula Used:
We will use the following formulas:
1.The expression of the form $\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {f\left( x \right)g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {f\left( x \right)} \right] \times \mathop {\lim }\limits_{x \to 0} \left[ {g\left( x \right)} \right]$ can be written as by splitting the limits.
2.L’Hopital’s rule states that $\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$.
3.The derivative of the function of the form ${e^x}$, is given as ${e^x}$.
4.The derivative of a constant is always 0.
5.The derivative of the variable with respect to itself is 1.

Complete step-by-step answer:
We will simplify the value of the expression using rules of limits, and L’Hopital’s rule.
First, we will simplify using the rules of limits.
Rewriting the expression, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x + x - x}} - {e^{x + 0}}}}{{\tan x - x}}$
Therefore, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x}{e^{\tan x - x}} - {e^x}{e^0}}}{{\tan x - x}}$
Factoring out ${e^x}$ from the numerator, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x}\left( {{e^{\tan x - x}} - {e^0}} \right)}}{{\tan x - x}}$
Rewriting the expression, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \left( {{e^x} \times \dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right)$
The expression of the form $\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {f\left( x \right)g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {f\left( x \right)} \right] \times \mathop {\lim }\limits_{x \to 0} \left[ {g\left( x \right)} \right]$ can be written as by splitting the limits.
Therefore, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \left( {{e^x} \times \dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {{e^x}} \right) \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right)$
Substituting $\mathop {\lim }\limits_{x \to 0} \left( {{e^x} \times \dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {{e^x}} \right) \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right)$ in the equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \left( {{e^x}} \right) \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right)$
Substituting the limit in the first expression of the product, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = {e^0} \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - {e^0}}}{{\tan x - x}}} \right)$
The value of any number raised to the power 0 is 1.
Thus, we get
$\Rightarrow {e^0} = 1$
Substituting ${e^0} = 1$ in the equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = 1 \times \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - 1}}{{\tan x - x}}} \right)$
Multiplying the terms, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - 1}}{{\tan x - x}}} \right)$
Now, we will use substitution.
Let $\tan x - x = t$.
If $x$ approaches 0, then the tangent of $x$ also approaches 0.
Therefore, we get
If $x \to 0$, then $\tan x - x \to 0$
Thus, we get
If $x \to 0$, then $t \to 0$
Substituting the new limit in the equation $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\tan x - x}} - 1}}{{\tan x - x}}} \right)$, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right)$
Now, we will use L’Hopital’s rule to simplify the expression.
L’Hopital’s rule states that $\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$.
Applying the L’Hopital’s rule, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \dfrac{{\dfrac{d}{{dt}}\left( {{e^t} - 1} \right)}}{{\dfrac{d}{{dt}}\left( t \right)}}$
The numerator is the derivative of the difference of two functions.
Therefore, rewriting the expression, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \dfrac{{\dfrac{d}{{dt}}\left( {{e^t}} \right) - \dfrac{d}{{dt}}\left( 1 \right)}}{{\dfrac{d}{{dt}}\left( t \right)}}$
The derivative of the function of the form ${e^x}$, is given as ${e^x}$.
The derivative of a constant is always 0.
The derivative of the variable with respect to itself is 1.
Simplifying the equation, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{e^t} - 0}}{1}$
Therefore, we get
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \dfrac{{{e^t}}}{1}\\\ \Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{t \to 0} \left( {{e^t}} \right)\end{array}$
Substituting the limit in the first expression of the product, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = {e^0}$
Substituting ${e^0} = 1$ in the equation, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = 1$
Substituting $\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}}$ in the equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = 1$
Therefore, we get the value of the expression $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}}$ as 1.
Thus, the correct option is option (c).

Note: We rewrote $\dfrac{d}{{dt}}\left( {{e^t} - 1} \right)$ as $\dfrac{d}{{dt}}\left( {{e^t}} \right) - \dfrac{d}{{dt}}\left( 1 \right)$. This is because the derivative of the difference of more than one functions is equal to the difference of the derivative of the individual functions, that is $\dfrac{{d\left( {p\left( x \right) - q\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left( {p\left( x \right)} \right)}}{{dx}} - \dfrac{{d\left( {q\left( x \right)} \right)}}{{dx}}$.
L’Hopital’s rule is only applicable if the expression is indeterminate, that is of the form $\dfrac{0}{0}$.
We can observe that by substituting the limit in the expression $\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right)$, we get
$\Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \dfrac{{{e^0} - 1}}{0}$
Therefore, we get
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \dfrac{{1 - 1}}{0}\\\ \Rightarrow \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right) = \dfrac{0}{0}\end{array}$
Since $\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{e^t} - 1}}{t}} \right)$ is of the form $\dfrac{0}{0}$, we can apply the L’Hopital’s rule.