Question

The value of $\underset{x\to 0}{\mathop{\lim}}\, \dfrac{{{e}^{{}^{1}/{}_{x}}}-1}{{{e}^{{}^{1}/{}_{x}}}+1}$ is equal to

a. 1
b. -1
c. 0
d. Does not exist.

Answer 1

Hint: For solving the above problem we would adopt the two-way limit process. First, we try to find out the left-hand limit and then evaluate the right-hand limit. After getting a definite number for both of the limits we compare whether both the values are the same or different. If they are the same then the limit exists and has a definite value. If both of them are different, then the limit does not exist.

Complete step-by-step answer:
We can approach a given number ‘a’ either from its left-hand or right-hand side. So, there are two types of limits viz. left-hand limit and right-hand limit. Also, for some functions at a given point ‘a’ left-hand and right-hand limit are equal whereas for some functions these two limits are not equal.
If $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)$ i.e., (L.H.L at x=a) = (R.H.L at x=a), then we say that $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists.
Let $f(x)=\dfrac{{{e}^{{}^{1}/{}_{x}}}-1}{{{e}^{{}^{1}/{}_{x}}}+1}$.
Now, we try to evaluate the left-hand limit at x=0:
\[\underset{x\to 0}{\mathop{\lim }}\,\text{f(x) = }\underset{h\to 0}{\mathop{\lim }}\,f(0-h)\]
Now, replacing x as (0 – h) we get,
\[\begin{align}
  & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{{}^{-1}/{}_{h}}}-1}{{{e}^{{}^{-1}/{}_{h}}}+1} \\
 & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{{{e}^{{}^{1}/{}_{h}}}}-1}{\dfrac{1}{{{e}^{{}^{1}/{}_{h}}}}+1} \right) \\
 & \Rightarrow \left( \dfrac{0-1}{0+1} \right)=-1 \\
\end{align}\]
So, the left-hand limit is -1.
Now, we try to evaluate the right-hand limit at x=0:
\[\underset{x\to 0}{\mathop{\lim }}\,\text{f(x) = }\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\]
Now, replacing x as (0 + h) we get,
\[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{e}^{{}^{1}/{}_{h}}}-1}{{{e}^{{}^{1}/{}_{h}}}+1}\]
Now, taking ${{e}^{{}^{1}/{}_{h}}}$ common from numerator and denominator and cancelling both we get,
\[\begin{align}
  & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{1-\dfrac{1}{{{e}^{{}^{1}/{}_{h}}}}}{1+\dfrac{1}{{{e}^{{}^{1}/{}_{h}}}}} \right) \\
 & \Rightarrow \left( \dfrac{1-0}{1+0} \right)=1 \\
\end{align}\]
So, the right-hand limit is 1.
Clearly, the left-hand limit and right-hand limit are different. So, from the theory we conclude that limit does not exist.
Therefore, the correct option is (d).

Note: Students must obtain both the values of limits and then give the conclusive answer. Because, if we obtain one side limit then the answer would be incorrect. An analysis of both sides of limits, LHL and RHL is important because if the limit does not exist, then evaluating the limit becomes inherently wrong.