Question

Jump of discontinuity of f(x) = $\frac{(e^{tan x} - 1)}{(e^{tan x} + 1)}$ at x = $\frac{\pi}{2}$ is

Answer 1

2

Answer 2

Let the given function be $f(x) = \frac{(e^{\tan x} - 1)}{(e^{\tan x} + 1)}$. We need to find the jump of discontinuity at $x = \frac{\pi}{2}$.

The jump of discontinuity at a point $x=a$ is defined as the absolute difference between the right-hand limit and the left-hand limit at that point, i.e., $|\lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x)|$.

First, we calculate the left-hand limit at $x = \frac{\pi}{2}$: $LHL = \lim_{x \to (\pi/2)^-} f(x) = \lim_{x \to (\pi/2)^-} \frac{e^{\tan x} - 1}{e^{\tan x} + 1}$

As $x \to (\frac{\pi}{2})^-$, $\tan x \to +\infty$.

Let $y = \tan x$. As $x \to (\frac{\pi}{2})^-$, $y \to +\infty$.

$LHL = \lim_{y \to +\infty} \frac{e^y - 1}{e^y + 1}$

To evaluate this limit, divide the numerator and the denominator by $e^y$: $LHL = \lim_{y \to +\infty} \frac{1 - e^{-y}}{1 + e^{-y}}$

As $y \to +\infty$, $e^{-y} \to 0$.

$LHL = \frac{1 - 0}{1 + 0} = 1$.

Next, we calculate the right-hand limit at $x = \frac{\pi}{2}$: $RHL = \lim_{x \to (\pi/2)^+} f(x) = \lim_{x \to (\pi/2)^+} \frac{e^{\tan x} - 1}{e^{\tan x} + 1}$

As $x \to (\frac{\pi}{2})^+$, $\tan x \to -\infty$.

Let $y = \tan x$. As $x \to (\frac{\pi}{2})^+$, $y \to -\infty$.

$RHL = \lim_{y \to -\infty} \frac{e^y - 1}{e^y + 1}$

As $y \to -\infty$, $e^y \to 0$.

$RHL = \frac{0 - 1}{0 + 1} = -1$.

Since the left-hand limit ($1$) and the right-hand limit ($-1$) exist but are not equal, the function has a jump discontinuity at $x = \frac{\pi}{2}$.

The jump of discontinuity is the absolute difference between the right-hand limit and the left-hand limit: Jump $= |RHL - LHL| = |-1 - 1| = |-2| = 2$.