Question

${lim}_{x\to 2}\frac{x^2-x-2}{x^2-2x+sin(x-2)}$

Answer 1

${lim}_{x\to 2}\frac{x^2-x-2}{x^2-2x+sin(x-2)}$

$={lim}_{x\to 2}\frac{(x-2)(x+1)}{x(x-2)+sin(x-2)}$

Dividing numerator and denominator by (x - 2)

$={lim}_{x\to 2}\frac{(x+1)}{x+\frac{sin(x-2)}{(x-2)}}$

$={lim}_{x\to 2}(x+1)\left(\frac{1}{x+\frac{sin(x-2)}{x-2}}\right)$

$={lim}_{x\to 2}(x+1)\times \frac{1}{{lim}_{x\to 2}}\left(x\right)+{lim}_{x\to 2}sin\frac{(x-2)}{x-2}$

$=(2\times 1)\times \frac{1}{\left(2\right)+{lim}_{x-2\to 0}\frac{sin(x-2)}{(x-2)}}$

$[\because {lim}_{0\to 0}\frac{sin\theta }{\theta }=1]$

$=3\times \frac{1}{2+1}$

$=1$