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ollegr [7]
3 years ago
6

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

Mathematics
1 answer:
Licemer1 [7]3 years ago
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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