Question

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

Answer 1

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

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

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

Dividing numerator and denominator by (x + 1)

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

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

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

$=\frac{(-1-2)}{(-1)+1}$ $[\because {lim}_{0\to 0}\frac{sin\theta }{\theta }=1]$

$=\frac{1}{0}$ $[\because \frac{1}{0}=\infty ]$

$=\infty$