Question

limit x - > 0, (tan^2x)/x Find the limit, I need some help

Answer 1

where do you think you shall begin

Answer 2

use l'hospitals rule

Answer 3

Well, I tried using that, But I usually get stuck at this part:

Answer 4

thats correct, now separate one sin x from sin^2 x and bring it about x, to get
sin x/x

Answer 5

\[\lim_{x \rightarrow 0}\frac{\sin x}{x}=1\]
someone correct me if im wrong

Answer 6

u are right.

Answer 7

\[\lim_{x \rightarrow 0} \frac{\tan^2x}{x} \]
\[\lim_{x \rightarrow 0} \frac{1}{x}*{\tan^2x} \]
\[\lim_{x \rightarrow 0} \frac{1}{x}*\frac{\sin^2x}{\cos^2x} \]
\[\lim_{x \rightarrow 0} \frac{\sin x}{x}*\frac{\sin x}{\cos^2x}=\lim_{x \rightarrow 0} \frac{\sin x}{x}*\lim_{x \rightarrow 0} \frac{\sin x}{\cos^2x}\]

Answer 8

\[\lim_{x \rightarrow 0} \frac{\sin x}{x}=1\]
please dont ask me to prove this, its just something you learn and will have to memorize

all you really need to do is solve
\[\lim_{x \rightarrow 0} \frac{\sin x}{\cos^2x}\]

Answer 9

if you're still confused, remember that
sin(0)=0 & cos(0)=1

Answer 10

Okay, I got all that. So, It'll be 0/1? =0?

Answer 11

yup

Answer 12

Thank you guys for the help :D

Answer 13

no prob