Limites

Creio que seja isso.

[latex]\\\mathrm{\lim_{x\to -1^-}\frac{x^2+x}{x+1}=\lim_{x\to -1^-}\frac{x(x+1)}{x+1}=\lim_{x\to -1^-}x=-1}\\\\\mathrm{\lim_{x\to -1^+}\frac{x^2+x}{x+1}=\lim_{x\to -1^+}\frac{x(x+1)}{x+1}=\lim_{x\to -1^+}x=-1}\\\\\mathrm{\lim_{x\to -1^-}\frac{x^2+x}{x+1}=\lim_{x\to -1^+}\frac{x^2+x}{x+1}\ \therefore \ \lim_{x\to -1}\frac{x^2+x}{x+1}=-1}\\\\\mathrm{f(x)\ \acute{e}\ continua\ se\ \lim_{x\to -1}\frac{x^2+x}{x+1}=f(-1),\ mas\ }\\\\\mathrm{f(-1)=2\neq \lim_{x\to -1}\frac{x^2+x}{x+1}=-1\ \therefore \ f(x)\ n\tilde{a}o\ \acute{e}\ continua\ em\ x=-1}\\\\\mathrm{Para\ x=0:\ f(0)=\frac{x^2+x}{x+1}=x=0}\\\\\mathrm{\lim_{x\to 0^-}\frac{x^2+x}{x+1}=\lim_{x\to 0^+}\frac{x^2+x}{x+1}=0\ \therefore \ \lim_{x\to 0}\frac{x^2+x}{x+1}=0}\\\\\mathrm{\lim_{x\to 0}\frac{x^2+x}{x+1}=f(0)=0\ \therefore \ f(x)\ \acute{e}\ continua\ em\ x=0}[/latex]