Answer
$\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)=1$
Work Step by Step
$$\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)$$
To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal.
$$\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)=\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big[\frac{2x+5}{x(x+1)}\Big]$$ $$\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)=\lim_{x\to-2^+}\Big(\frac{1}{x+1}\Big)\Big(\frac{2x+5}{x+1}\Big)=\lim_{x\to-2^+}\frac{2x+5}{(x+1)^2}$$ $$\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)=\frac{2\times(-2)+5}{(-2+1)^2}=\frac{1}{(-1)^2}=1$$
Therefore, $\lim_{x\to-2^+}\Big(\frac{x}{x+1}\Big)\Big(\frac{2x+5}{x^2+x}\Big)=1$
