euniceolorunsho euniceolorunsho

04-04-2019
Mathematics

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evaluate the limit lim x—->0 sin(2x)/x^2+x

Respuesta :

LammettHash LammettHash

08-04-2019

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]

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