Question

Find the limit of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$.
A. 1
B. -1
C. 0
D. does not exist

Answer 1

Hint: We first need to find if the limit even exists or not. If the limit exists then we find its value. For the function to have a limit value its left-hand limit and right-hand limit both have to be equal. We assume small value $h\to {{0}^{+}}$ and find both $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{-h}}}-1}{{{e}^{\dfrac{1}{-h}}}+1}$ and $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}$. The values don’t match and that’s why the limit of the $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$ doesn’t exist.

Complete step-by-step solution
To find the limit of the function $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$, we first need to check if the limit exists or not.
The limit of $f\left( x \right)$ exists at a when $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)$. This means the left-hand limit and right-hand limit both are equal then only the actual limit exists.
Let us check the left-hand limit of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$. We are finding the limit at 0. ${{0}^{-}}$ defines the small value at the negative end. Let’s take it as $\left( 0-h \right)=-h$ where $h\to {{0}^{+}}$. We find the limit value as $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{-h}}}-1}{{{e}^{\dfrac{1}{-h}}}+1}$. We find individual values.
The value of “h” is very small, and the inverse will be very big. So, ${{e}^{\dfrac{1}{h}}}\to \infty $.
Then we are taking negative of it which means ${{e}^{\dfrac{1}{-h}}}=\dfrac{1}{{{e}^{\dfrac{1}{h}}}}\to 0$.
Now we place the values in $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{-h}}}-1}{{{e}^{\dfrac{1}{-h}}}+1}$.
$\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{-h}}}-1}{{{e}^{\dfrac{1}{-h}}}+1}=\dfrac{-1}{1}=-1$.
Now we check the right-hand limit of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$. We are finding the limit at 0. ${{0}^{+}}$ defines the small value at the positive end. Let’s take it as $\left( 0+h \right)=h$ where $h\to {{0}^{+}}$. We find the limit value as $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}$. We find individual values.
The value of “h” is very small, and the inverse will be very big. So, ${{e}^{\dfrac{1}{h}}}\to \infty $.
Now we place the values in $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}$. In this both denominator and numerator the total values of ${{e}^{\dfrac{1}{h}}}-1$ and ${{e}^{\dfrac{1}{h}}}+1$ tends to infinity. So, if $h\to {{0}^{+}}$ then ${{e}^{\dfrac{1}{h}}}-1\to \infty ,{{e}^{\dfrac{1}{h}}}+1\to \infty $.
$\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}=\dfrac{\infty }{\infty }=1$.
So, we can see the two-limit value doesn’t match. $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)\ne \displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)$.
So, the limit $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$ doesn’t exist. The correct option is D.

Note: If the notion $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}=\dfrac{\infty }{\infty }=1$ is tough to understand then we can use L’hospital rule where in case of $\dfrac{\infty }{\infty },\dfrac{0}{0}$ form we use differentiation of both numerator and denominator.
So, $\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}-1}{{{e}^{\dfrac{1}{h}}}+1}=\displaystyle \lim_{h\to 0}\dfrac{{{e}^{\dfrac{1}{h}}}}{{{e}^{\dfrac{1}{h}}}}\times \dfrac{\dfrac{-1}{{{h}^{2}}}}{\dfrac{-1}{{{h}^{2}}}}=\displaystyle \lim_{h\to 0}1=1$.
This way we can express the limit and find that the limit of $\displaystyle \lim_{x \to 0}\dfrac{{{e}^{\dfrac{1}{x}}}-1}{{{e}^{\dfrac{1}{x}}}+1}$ doesn’t exist.