Question

If $f\left( x \right)=\dfrac{1+{{e}^{\dfrac{1}{x}}}}{1-{{e}^{\dfrac{1}{x}}}}\text{ }\left( x\ne 0 \right)\text{ and f}\left( 0 \right)=1$ , the f(x) is
(a) Left continuous at x=0
(b) Right continuous at x=0
(c) Continuous at x=0
(d) None

Answer 1

Hint: Check the continuity of the function f(x) at x=0, and if it is not continuous check for the one sided continuity of the function at x=0. For finding the RHL, try to eliminate the ${{e}^{\dfrac{1}{x}}}$ from the denominator and numerator, while for finding the LHL, directly put the limit.

Complete step-by-step answer:
For f(x) to be continuous at x=a, the value of the function at x=a must be equal to $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)\text{ and }\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)$ .
Therefore, checking the continuity of the function $f\left( x \right)=\dfrac{1+{{e}^{\dfrac{1}{x}}}}{1-{{e}^{\dfrac{1}{x}}}}\text{ }\left( x\ne 0 \right)\text{ and f}\left( 0 \right)=1$ .
First, finding the right-hand limit of the given function.
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1+{{e}^{\dfrac{1}{x}}}}{1-{{e}^{\dfrac{1}{x}}}}$
Now, if we take ${{e}^{\dfrac{1}{x}}}$ from both, the denominator and the numerator of the expression, we get
$\underset{x\to {{0}^{+}}}{\mathop{=\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}\left( 1+{{e}^{-\dfrac{1}{x}}} \right)}{{{e}^{\dfrac{1}{x}}}\left( {{e}^{-\dfrac{1}{x}}}-1 \right)}$
Now we know that as x approaches to 0 from the right side, the value of $-\dfrac{1}{x}$ approaches $-\infty$ , hence the value of ${{e}^{-\dfrac{1}{x}}}$ tends to be zero.
$\therefore \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{e}^{\dfrac{1}{x}}}\left( 1+{{e}^{-\dfrac{1}{x}}} \right)}{{{e}^{\dfrac{1}{x}}}\left( {{e}^{-\dfrac{1}{x}}}-1 \right)}=\dfrac{1+0}{0-1}=-1$
Next, finding the left hand limit.
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{1+{{e}^{\dfrac{1}{x}}}}{1-{{e}^{\dfrac{1}{x}}}}$
Now we know that as x approaches to 0 from the left side, the value of $\dfrac{1}{x}$ approaches $-\infty$ , hence the value of ${{e}^{\dfrac{1}{x}}}$ tends to be zero.
$\therefore \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{1+{{e}^{\dfrac{1}{x}}}}{1-{{e}^{\dfrac{1}{x}}}}=\dfrac{1+0}{1-0}=1$

As we can see, the left-hand limit of the function is equal to the value of the limit at zero, so the function is left continuous at x=0. Hence, the answer to the above question is option (a).

Note: Whenever you come across the forms $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ always give a try to L-hospital’s rule as this may give you the answer in the fastest possible manner. Also, keep a habit of checking the indeterminate forms of the expressions before starting with the questions of limit as this helps you to select the shortest possible way to reach the answer.