Question

Cho biết \(\lim\limits_{x\rightarrow0}\dfrac{sinax}{ax}=1\left(a\ne0\right)\). Tìm \(\lim\limits_{x\rightarrow0}\dfrac{1-cos2017x}{x^2}\)

Answer 1

\(\lim\limits_{x\rightarrow0}\dfrac{\sin ax}{ax}=1\Rightarrow\sin ax\sim ax\Leftrightarrow\sin^2ax\sim\left(ax\right)^2\)

\(1-\cos x=1-\cos2.\dfrac{x}{2}=2\sin^2\dfrac{x}{2}\sim2.\left(\dfrac{x}{2}\right)^2=\dfrac{x^2}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{1-\cos2017x}{x^2}\)

Ta co khi \(x\rightarrow0:1-\cos2017x\sim\dfrac{\left(2017x\right)^2}{2}=\dfrac{2017^2x^2}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{1-\cos2017x}{x^2}=\lim\limits_{x\rightarrow0}\dfrac{2017^2x^2}{2x^2}=\dfrac{2017^2}{2}\)