Question

Find if the limit of function exists as x tends to zero.
$f\left( x \right)=\left\{ \begin{align}
  & \dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\ne 0 \\
 & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=0 \\
\end{align} \right.$

Answer 1

Hint: In order to solve this problem, we need to find the left-hand limit and the right-hand limit of the function. If the left-hand limit is equal to the right-hand limit then the limit of the function exists and if the left-hand limit and right-hand limit does not exist then the limit does not exist.

Complete step-by-step answer:
We have given a function and we need to find whether the limit exists if it approaches zero.
To find that we need to approach the function from a negative direction and positive direction and see whether its value is the same or not.
If the value is the same then the limit exists as the function tends to zero and when the value is not the same then the limit does not exist.
Let's take the negative limit first
We need to find the limit of this quantity $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$
Substituting the value of ${{0}^{-}}$ in the limit we get,
$\underset{x\to {{0}^{-}}}{\mathop{\lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{{{e}^{\dfrac{1}{{{0}^{-}}}}}-1}{{{e}^{\dfrac{1}{{{0}^{-}}}}}+1}$
$\dfrac{1}{{{0}^{=}}}$ tends to negative infinity,
Therefore, substituting we get,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{{{e}^{-\infty }}-1}{{{e}^{-\infty }}+1}$
We must know the ${{e}^{-\infty }}$ is always 0.
Therefore substituting we get,
$\underset{x\to{{0}^{-}}}{\mathop{\lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{0-1}{0+1}=-1$
Therefore, the left-hand limit is -1.
Now let's find the right-hand limit.
We need to find the limit of this quantity $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$
Substituting the value of ${{0}^{-}}$ in the limit we get,
$\underset{x\to{{0}^{+}}}{\mathop{\lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{{{e}^{\dfrac{1}{{{0}^{+}}}}}-1}{{{e}^{\dfrac{1}{{{0}^{+}}}}}+1}$
$\dfrac{1}{{{0}^{+}}}$ tends to positive infinity,
Therefore, substituting we get,
$\underset{x\to{{0}^{+}}}{\mathop{\lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{{{e}^{\infty }}-1}{{{e}^{\infty }}+1}$
We must know the ${{e}^{\infty }}$ is always infinite.
Therefore substituting we get,
$\underset{x\to{{0}^{-}}}{\mathop{\lim}}\,\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}=\dfrac{\infty -1}{\infty +1}=\dfrac{\infty }{\infty }$
But, $\dfrac{\infty }{\infty }$ is an indeterminate form. So we need to find by l’hopital rule.
In this when the form of limit is indeterminate then we need to differentiate the numerator and denominator separately.
Therefore, by doing this we get,
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,=\dfrac{\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}}-1 \right)}{\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}}+1 \right)}=\dfrac{{{e}^{\dfrac{1}{x}}}\times -\dfrac{1}{{{x}^{2}}}}{{{e}^{\dfrac{1}{x}}}\times -\dfrac{1}{{{x}^{2}}}}=1$
Therefore, the right-hand limit is 1.
Now by comparing the left-hand limit and right-hand limit we get,
The left-hand limit is not equal to the right-hand limit,
Hence the limit to the function does not exist as the function tends to zero.
Hence the limit to the function does not exist as the function tends to zero.

Note: We need to know the values of ${{e}^{-\infty }}$ and ${{e}^{\infty }}$ . The value of ${{e}^{-\infty }}$ is always 0 and the value of ${{e}^{\infty }}$ is infinite. When we write the term ${{0}^{-}}$ we mean that we are just at the negative of zero. And when we write ${{0}^{+}}$ we are just to the positive side of the zero. Therefore, by calculating the value just to either side we can find whether the limit exists or not.