Question

Show that $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}}} \right)$ does not exist?

Answer 1

Limit is a value for a function that approaches some value. It is an important part of calculus that is used to define continuity, derivatives and integrals. Limit is solved by taking the Left-hand limit and right-hand limit. Solving them separately, if they are equal then the limit exists otherwise do not exist.

Complete answer: We are given a function, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}}} \right)$. To check whether the limit of the function exists or not, we need to check for the LHL (Left Hand Limit) and RHL (Right Hand Limit).
So, starting with the Left-Hand-Limit, that is the left side for the value $x \to 0$, that is written as $x \to {0^ - }$:
LHL:
$\mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{{e^{\dfrac{1}{{ - 0}}}} - 1}}{{{e^{\dfrac{1}{{ - 0}}}} + 1}}} \right)$
Since, we know that $\dfrac{1}{0} = \infty$, so substituting this in the above value, we get:
$\mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{{e^{\dfrac{1}{{ - 0}}}} - 1}}{{{e^{\dfrac{1}{{ - 0}}}} + 1}}} \right) = \left( {\dfrac{{{e^{ - \infty }} - 1}}{{{e^{ - \infty }} + 1}}} \right)$
Now, we know that ${e^\infty } = \infty$ but for ${e^{ - \infty }} = \dfrac{1}{{{e^\infty }}} = \dfrac{1}{\infty } = 0$.
So, substituting this value in the above limit, we get:
$\Rightarrow \left( {\dfrac{{{e^{ - \infty }} - 1}}{{{e^{ - \infty }} + 1}}} \right) = \left( {\dfrac{{0 - 1}}{{0 + 1}}} \right) = - 1$
Therefore, the LHL for the $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}}} \right)$ is $- 1$.
Now, for RHL:
Now, for the Right side of the value $x \to 0$, that is written as $x \to {0^ + }$:
$\mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{{{e^{\dfrac{1}{{ + 0}}}} - 1}}{{{e^{\dfrac{1}{{ + 0}}}} + 1}}} \right)$
Since, we know that $\dfrac{1}{0} = \infty$, so substituting this in the above value, we get:
$\mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{{e^{\dfrac{1}{0}}} - 1}}{{{e^{\dfrac{1}{0}}} + 1}}} \right) = \left( {\dfrac{{{e^\infty } - 1}}{{{e^\infty } + 1}}} \right)$
Taking ${e^\infty }$ common, we get:
$\Rightarrow \dfrac{{{e^\infty }\left( {1 - \dfrac{1}{{{e^\infty }}}} \right)}}{{{e^\infty }\left( {1 + \dfrac{1}{{{e^\infty }}}} \right)}}$
Cancelling out the term ${e^\infty }$:
$\Rightarrow \dfrac{{\left( {1 - \dfrac{1}{{{e^\infty }}}} \right)}}{{\left( {1 + \dfrac{1}{{{e^\infty }}}} \right)}}$
Now, we know that ${e^\infty } = \infty$ but for $\dfrac{1}{{{e^\infty }}} = \dfrac{1}{\infty } = 0$.
So, substituting this value in the above limit, we get:
$\Rightarrow \dfrac{{\left( {1 - \dfrac{1}{{{e^\infty }}}} \right)}}{{\left( {1 + \dfrac{1}{{{e^\infty }}}} \right)}} = \dfrac{{1 - 0}}{{1 + 0}} = 1$
Therefore, the RHL for the $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}}} \right)$ is $1$.
Since, we can see that the LHL and RHL are not equal for the limit, that means the limit for the function $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{e^{\dfrac{1}{x}}} - 1}}{{{e^{\dfrac{1}{x}}} + 1}}} \right)$ does not exist.

Note:
It’s important to always check for both the left and right side of the limit for the function, in order to say whether a limit exists or not.
There are certain rules or properties that are followed for limits known as the “Law of Limits”- They are as follows- The Notation of a Limit, The sum Rule, The Extended Sum Rule, The Constant Function Rule, The Constant Function Rule, The Constant Multiple Rule and many more.