# Limit of a Trigonometric Function

1. Oct 9, 2005

### dekoi

Question:

lim(x->0) for (tanx - sinx) / (sinx)^2

This is what I got:

= (sinx-sinxcosx) / (cosx)(sinx)^2
= (sinx)(1-cosx) / (sinx)(sinx)(cosx)
= (1 - cosx) / (sinx)(cosx)

However, I can't figure out what to do from this step, as the limit still equals 0/0 at this stage.

2. Oct 9, 2005

### mezarashi

A 0/0 answer is the right prerequisite for using L'Hopital's rule.

That is, the limit of any such function f(x) = h(x)/g(x), is h'(x)/g'(x).
Try differentiating the top and bottom separately and plug in the numbers again.

3. Oct 9, 2005

### dekoi

I haven't learned that yet.

We are expected to solve the problem with using only limit laws and the fact that the lim (x->0) for sinx / x equals 1.

4. Oct 9, 2005

??? .

5. Oct 9, 2005

### mezarashi

Then you need to approach the problem differently. It is purely an algebriac/trigonometric problem. The strategy is to rid the denominator of any possible 0 terms (i.e. sin x).

Edit - Here, try this:
$$\frac{\tan x - \sin x}{sin^2x}$$

$$\frac{\frac{\sin x - \sin x \cos x}{\cos x}}{1 - \cos^2 x}$$

Keep in mind:
$$(1-a^2) = (1-a)(1+a)$$

Last edited: Oct 9, 2005
6. Oct 10, 2005

### HallsofIvy

Do you know that $$lim_{x\rightarrow0}\frac{sin x}{x}= 1$$?
Do you know that $$lim_{x\rightarrow0}\frac{1- cos x}{x}= 0$$?

Can you figure out how to write $$\frac{1-cos x}{sin x}$$ in terms of $$\frac{sin x}{x}$$ and $$\frac{1- cos x}{x}$$?

7. Oct 10, 2005

### ivybond

(1 - cosx) / (sinx)(cosx) *
(1+cosx) / (1+cosx) = ...

Or:
(tanx - sinx) / (sinx)^2 =
tanx (1 - cosx) / (1 - (cosx)^2) = ...
(look up Mezarashi's hint)

Last edited: Oct 11, 2005
8. Nov 20, 2011

lim x->0 of (tanx - sinx)/(sinx)^2

lim x->0 of (tanx)/(sinx)^2 - (sinx)/(sinx)^2

lim x->0 of (sinx)/(cosx(sinx)^2) - (1/sinx)

lim x->0 (1/cosxsinx) - (1/sinx)

lim x->0 (1-cosx)/(cosxsinx)

lim x->0 (-(cosx-1)/x) / (cosxsinx)/(x))

lim x->0 (-0)/ (1(1)) = 0

so the final answer is 0