Limits of Trig Functions

Hello all. I missed a class in calculus so I didn't get the notes on this so if anyone could explain this question for me, it would be much appreciated.

$$\lim_{x \rightarrow 0} \frac {tanx}{4x}$$
$$= \frac {sinx}{cos4x} ?$$

Not really too sure if I manipulated the equation right. Any hints for the next step? Thanks.

You can't do that!!!!!

Use l'hopitals rule.

$$\lim_{x \rightarrow 0} \frac {tanx}{4x} =\lim_{x \rightarrow 0} \frac {secxtanx}{4}$$

Sorry, I haven't learned l'hopitals rule yet and we're not suppose to use it for these questions.

expand tanx in taylor series, and do what you should do....

This is going to sound really pathetic but no, we haven't the taylor series either.

do you know the fact that
$$\lim_{x \rightarrow 0} \frac {sinx}{x} =1$$
if yes, you should start from here

dextercioby
Homework Helper
I think Taylor series is taught way after l'Hôpital's rule,don't u think so? Daniel.

vincentchan said:
do you know the fact that
$$\lim_{x \rightarrow 0} \frac {sinx}{x} =1$$
if yes, you should start from here

That I do know.

dextercioby
Homework Helper
What about "tangent's" definition...?And the limit of cosine as its argument goes to 0 ?

Daniel.

erik05 said:
Hello all. I missed a class in calculus so I didn't get the notes on this so if anyone could explain this question for me, it would be much appreciated.

$$\lim_{x \rightarrow 0} \frac {tanx}{4x}$$
$$= \frac {sinx}{cos4x} ?$$

Not really too sure if I manipulated the equation right. Any hints for the next step? Thanks.

= 1/4*(sin[x]/x)*(sec[x])

Ah...I got it. Thanks all.

Limit Laminate...

Solution:
$$\boxed{\lim_{x \rightarrow 0} \frac {\tan x}{4x} = \frac{1}{4}}$$

Last edited:
dextercioby
How fancy that \boxed{...},too bad u don't know "\tan"... P.S.BTW,I've searched Mathworld and A & S,couln't find this $tanx$ function... P.P.S.Neither $sinx$,nor $secxtanx$,but i found $\mbox{sinc}\ x$... :surprised
P.P.P.S.You edited... 