Question

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

Answer 1

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.

Complete step-by-step answer:
As mentioned in the question, we have to evaluate the limit as given in the question.
Now, as both the numerator as well as the denominator is forming the indeterminate form, hence, we can apply L’Hospital rule as given in the hint that is as follows

& \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{aligned}$$ Hence, the limit of the given function is equal to 2. Note: The students can make an error if finding the value of k if they don’t know the basic definition of L’Hospital rule and where can we apply it which is also given in the hint as follows The formula that will be used in the solution is as follows $$\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.