Question

Evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{3x}}}}{{\sqrt {1 + x} - \sqrt {1 - x} }}$

Answer 1

We can apply the limit directly to check whether the limit becomes $\dfrac{0}{0}$ . If the limit becomes $\dfrac{0}{0}$ , we can apply L’ Hospital’s rule which is given by $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ . We can find the derivatives of the functions separately. Then we can substitute in the equation. Then we can apply the limits to get the value of the required limit.

Complete step-by-step answer:
We need to find the value of $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{3x}}}}{{\sqrt {1 + x} - \sqrt {1 - x} }}$
Let $I = \mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{3x}}}}{{\sqrt {1 + x} - \sqrt {1 - x} }}$
We can directly apply the limits. For that we can substitute $x = 0$ .
$\Rightarrow I = \dfrac{{{e^{5 \times 0}} - {e^{3 \times 0}}}}{{\sqrt {1 + 0} - \sqrt {1 - 0} }}$
We know that, ${e^0} = 1$ . On applying this relation, we get,
$\Rightarrow I = \dfrac{{1 - 1}}{{1 - 1}}$
On further simplification, we get,
$\Rightarrow I = \dfrac{0}{0}$
So, we can apply L’ Hospital’s rule. According to this rule, if $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$ ,then $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ .
Here $f\left( x \right) = {e^{5x}} - {e^{3x}}$ and $g\left( x \right) = \sqrt {1 + x} - \sqrt {1 + x}$
Now we can find $f'\left( x \right)$ and $g'\left( x \right)$
$f'\left( x \right) = \dfrac{d}{{dx}}{e^{5x}} - {e^{3x}}$
We know that, $\dfrac{d}{{dx}}{e^{ax}} = a{e^{ax}}$
$\Rightarrow f'\left( x \right) = 5{e^{5x}} - 3{e^{3x}}$
$g'\left( x \right)$ is given by,
$g'\left( x \right) = \dfrac{d}{{dx}}\left( {\sqrt {1 + x} - \sqrt {1 + x} } \right)$
We know that $\dfrac{d}{{dx}}\sqrt x = \dfrac{1}{{2\sqrt x }}$ and by applying chain rule of differentiation, we get
$\Rightarrow g'\left( x \right) = \dfrac{1}{{2\sqrt {1 + x} }}\left( 1 \right) - \dfrac{1}{{2\sqrt {1 - x} }}\left( { - 1} \right)$
$\Rightarrow g'\left( x \right) = \dfrac{1}{{2\sqrt {1 + x} }} + \dfrac{1}{{2\sqrt {1 - x} }}$
By L’ Hospital’s Rule, $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ , we can write the given limit as,
$\Rightarrow I = \mathop {\lim }\limits_{x \to 0} \dfrac{{5{e^{5x}} - 3{e^{3x}}}}{{\dfrac{1}{{2\sqrt {1 + x} }} + \dfrac{1}{{2\sqrt {1 - x} }}}}$
On applying the limits, we get,
$\Rightarrow I = \dfrac{{5{e^{5 \times 0}} - 3{e^{3 \times 0}}}}{{\dfrac{1}{{2\sqrt {1 + 0} }} + \dfrac{1}{{2\sqrt {1 - 0} }}}}$
We know that, ${e^0} = 1$ . On applying this relation, we get,
$\Rightarrow I = \dfrac{{5 - 3}}{{\dfrac{1}{2} + \dfrac{1}{2}}}$
On adding the terms in the denominator, we get,
$\Rightarrow I = \dfrac{2}{1}$
On further simplification, we get,
$\Rightarrow I = 2$
So the required value of the limit is 2.
Therefore, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{5x}} - {e^{3x}}}}{{\sqrt {1 + x} - \sqrt {1 - x} }} = 2$ .

Note: When we get an expression to find the limit, we can directly apply the limits to check whether it is defined. Then we can use different methods to simplify the expression and then apply the limits to get the required value of limit. We can use only the L’ Hospital’s rule $\mathop {\lim }\limits_{x \to \infty } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ , only when the limit tends to $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ .
Similarly, we can use the relation $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$ when $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {1^\infty }$.