# Trigonometric limits - question involving tan.

• October 6th 2009, 11:50 AM
Kakariki
Trigonometric limits - question involving tan.
I am having trouble answering this question:

"Using $tanx = \frac{sinx}{cosx}$ evaluate *limit as x approaches 0* $\frac{tanx - sinx}{xcosx}$"

My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

My solution so far:
note:*the "limit as x approaches 0" I don't know how to input on latex, so pretend it is there before all these.*

$= \frac{\frac{sinx}{cosx} - sinx}{xcosx}$
$= \frac{\frac{sinx - sinxcosx}{cosx}}{xcosx}$
$= \frac {sinx - sinx}{xcosx}$

Don't know what to do from here. I know the answer is zero, but I do not know how to get there from what I have. Help please.
• October 6th 2009, 11:59 AM
josipive
• October 6th 2009, 12:14 PM
Soroban
Hello, Kakariki!

We're expected to know: . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

Quote:

I am having trouble answering this question:

Using $\tan x \:=\: \frac{\sin x}{\cos x}$, evaluate: . $\lim_{x\to0}\frac{\tan x - \sin x}{x\cos x}$

My main problem, I think, is getting rid of the x on the bottom, I do not know how to go about doing it.

My solution so far:

$\lim_{x\to0} \frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}$

. . $=\;\lim_{x\to0}\frac{\frac{\sin x - \sin x\cos x}{\cos x}}{x\cos x}$

. . $= \;\lim_{x\to0}\frac {\sin x - \sin x}{x\cos x}$ . . . . wrong!

We have: . $\frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}$

Multiply by $\frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\frac{\sin x}{\cos x} - \sin x\right)}{\cos x(x\cos x)} \;=\;\frac{\sin x - \sin x\cos x}{x\cos^2\!x} \;=\;\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}$

$\text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left( \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}$

. . . . . . . . . $= \;1\cdot0 \;=\;0$

• October 6th 2009, 12:32 PM
Kakariki
Quote:

Originally Posted by Soroban
Hello, Kakariki!

We're expected to know: . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1$

We have: . $\frac{\tan x - \sin x}{x\cos x} \;=\;\frac{\frac{\sin x}{\cos x} - \sin x}{x\cos x}$

Multiply by $\frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\frac{\sin x}{\cos x} - \sin x\right)}{\cos x(x\cos x)} \;=\;\frac{\sin x - \sin x\cos x}{x\cos^2\!x} \;=\;\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}$

$\text{Take the limit: }\;\lim_{x\to0}\bigg[\frac{\sin x}{x}\cdot\frac{1-\cos x}{\cos^2\!x}\bigg] \;=\;\underbrace{\lim_{x\to0}\left(\frac{\sin x}{x}\right)}_1\cdot\underbrace{\lim_{x\to0}\left( \frac{1-\cos x}{\cos^2\!x}\right)}_{\frac{1-1}{1^2}}$

. . . . . . . . . $= \;1\cdot0 \;=\;0$

Okay, that makes sense. How did you know to multiply by $\frac{cosx}{cosx}$ though? Is it because you can't have a 0 in the denominator, but you can in the numerator?