Respuesta :

LammettHash

Recall that

[tex]\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=1[/tex]

for [tex]a\neq0[/tex].

We have

[tex]\displaystyle\lim_{x\to0}\frac{\sin2x}{x^2+x}=\lim_{x\to0}\frac{\sin2x}{2x}\cdot\lim_{x\to0}\frac2{x+1}=2[/tex]