Question regarding limits to see if my procedure of getting the answer is right!!!!

1 answer:

7 0

$\displaystyle\lim_{x\to0}\frac{\tan x^2}x=\lim_{x\to0}x\frac{\tan x^2}{x^2}=\lim_{x\to0}\frac x{\cos x^2}\times\lim_{x\to0}\frac{\sin x^2}{x^2}$

Recall that

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

and so replacing $x$ with $x^2$ , you get that the second limit is 1. Meanwhile,

$\displaystyle\lim_{x\to0}\dfrac x{\cos x^2}=\dfrac01=0$

so the limit is 0.

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