Is my solution to the mentioned problem correct ?
If $y=\frac{1}{(x+a)} \;\;\;\; (for\; x \neq -a)\tag{1}$
$\Delta y = \frac{1}{(x+\Delta x + a)} - \frac{1}{(x+a)}$
$\Delta y = (x+\Delta x +a)^{-1} - (x+a)^{-1}$
$\Delta y = \frac{-\Delta x}{(x+a)(x+\Delta x+a)}$
$\frac {\Delta y}{\Delta x} = \frac{-1}{(x+a)(x+\Delta x+a)} \tag{2}$
Differentiating Eq. 1 w.r.t x,
$\frac{dy}{dx}=(-1)(x+a)^{-2} \tag{3}$
As, $\epsilon= \frac{\Delta y}{\Delta x}-\frac{dy}{dx} \tag{4}$
Hence putting values of Eq. 2 & 3 in 4,
$\epsilon= \frac{-1}{(x+a)(x+\Delta x+a)} + \frac{1}{(x+a)^{2}}$
$\epsilon= \frac{1}{(x+a)}\left[\frac{-1}{(x+\Delta x+a)} + \frac{1}{(x+a)} \right]$
Finally,
$\epsilon= \frac{\Delta x}{(x+a)^{2}(x+\Delta x+a)} \;\;\;\; (for\; x \neq -a) \tag{5}$
Hence as $\Delta x \to 0, \epsilon \to 0$
The answer of the question is written in the book as below (which doesn't match with my answer):
$\epsilon= \frac{(x+a)(\Delta x)}{(x+a)(x+\Delta x+a)} \tag{6}$