QUESTION

# Find the value of the following: $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-{{e}^{-x}}}{x}$

Hint: The formula that will be used in the solution is as follows-
$\underset{x\to t}{\mathop{\lim }}\,f(x)=\underset{x\to t}{\mathop{\lim }}\,\dfrac{g(x)}{h(x)}=\dfrac{{g}'(t)}{{h}'(t)}$
The above formula is known as L’Hospital rule and it is used to find the limit of a particular function.
This formula or property is only applicable if the numerator as well as the denominator is forming an indeterminate form.

\begin{align} & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{x}}-{{e}^{-x}}}{x} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\left( {{e}^{x}}-{{e}^{-x}} \right)}^{\prime }}}{{{\left( x \right)}^{\prime }}} \\ & =\dfrac{{{e}^{0}}+{{e}^{-0}}}{1} \\ & =1+1 \\ & =2 \\ \end{align}
$\underset{x\to t}{\mathop{\lim }}\,f(x)=\underset{x\to t}{\mathop{\lim }}\,\dfrac{g(x)}{h(x)}=\dfrac{{g}'(t)}{{h}'(t)}$