Question

im having trouble finding the limit as x goes to 0 of x(1-cosx)/tan^3(x).

Answer 1

any luck solving this? im stumped

Answer 2

\[\lim_{x\to0}\frac{x(1-\cos x)}{\tan^3x}~~?\]

Answer 3

yes that!

Answer 4

Okay, well near \(x=0\), you have \(\tan x\approx x\), so you can rewrite to get
\[\lim_{x\to0}\frac{1-\cos x}{\tan^2x}\]
Follow so far?

Answer 5

is that a trig rule/identity thing?

Answer 6

Kind of. It's a trick for limits. If you're not comfortable with that, you can alternatively rewrite the tangent using its appropriate identity:
\[\lim_{x\to0}\frac{x(1-\cos x)}{\tan x\tan^2x}\]
Then some factoring/rearranging:
\[\lim_{x\to0}\frac{x}{\tan x}\cdot\frac{1-\cos x}{\sec^2x-1}\]

Answer 7

i like the trick for limits better. rather know that because im not very good with the trig identities.

Answer 8

Hmm, something tells me that trick won't work for this problem... Coming to a roadblock with it.

Answer 9

Are you familiar with the following limits:
\[\lim_{x\to0}\frac{\sin x}{x}=1~~\text{and}~~\lim_{x\to0}\frac{1-\cos x}{x}=0~~?\]

Answer 10

could you use the \[\tan ^{2}\] where its equal to \[\frac{ 1-\cos(2u) }{ 1+\cos(2u)}\]?
and yes i know those limits.

Answer 11

In that case, I'd prefer to use those rather than any unnecessary identities.

We left off with
\[\lim_{x\to0}\frac{1-\cos x}{\tan^2x}\]
Multiplying by \(\dfrac{x}{x}\)
\[\lim_{x\to0}\frac{1-\cos x}{x}\cdot\frac{x}{\tan^2x}\]

Answer 12

The limit can be split up:
\[\color{red}{\lim_{x\to0}\frac{1-\cos x}{x}}\cdot\lim_{x\to0}\frac{x}{\tan^2x}\]

Answer 13

Hmm, the answer is apparently not 0... Forget that then. I'll see what else can be done.

Answer 14

if you used the \[tanx \approx x\] could the bottom be \[x ^{3}\]. then x would cancel with \[x ^{3}\] and give the limit for \[\frac{ (1-cosx) }{ x^2 }\] so it would be 1/2.

Answer 15

the limit is for sure 1/2 because wolfram says so i just didnt know how to do it.

Answer 16

How about
\[\frac{1-\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}=\frac{\sin^2x}{x^2(1+\cos x)}\]
That should do the trick.