# Tricky Limits

1. Sep 7, 2006

### vgower

Limits

I'm doing some online practice work... and I can't find out how to get the following two limits.... I know they are not undefined.

The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.

The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

Any help would be appreciated.

2. Sep 7, 2006

3. Sep 7, 2006

### vgower

We're not doing derivitaves yet.. so it wouldnt make sense.. but I could try.

4. Sep 7, 2006

### vgower

Doesn't work.

5. Sep 7, 2006

### vgower

I'm doing some online practice work... and I can't find out how to get the following two limits.... I know they are not undefined.

The limit as x approaches 0 of sin^2(4x) / xtan(x) .. I tried to flip it... but I couldn't get it that way either.

The limit as x approaches 0 of (x + tan(4x)) / sin(9x)

Any help would be appreciated.

6. Sep 7, 2006

### Swapnil

I am getting 16 as my answer. I first broke $$\sin(4x)^2$$ into $$(2\sin(2x)*\cos(2x))^2$$ and then I further broke this down to $$16\sin(x)^2\cos(x)^2(\cos(x)^2-\sin(x)^2)^2$$. Then I didived it by x*tan(x) from which I got:

$$16*\frac{\sin(x)}{x}\cos(x)^3(\cos(x)^2-\sin(x)^2)$$

Then ... here is the muddy part ... I assumed that the limit of the products is the product of the limits. Then, since the limit of sinx(x)/x is 1, you end up with 16 as your answer after you evalute the rest of the limit at x=0.

7. Sep 7, 2006

### Hurkyl

Staff Emeritus
Then find a reason why you can assume that. It's not hard.

Incidentally, if you know (sin x)/x --> 1 as x --> 0, there's a much easier way to deal with sin (4x)...

8. Sep 7, 2006

### dmoravec

i also got 16 for the first one using L'Hopitals rule (although I had to go to second derivatives) but I just expanded sin^2(4x) as [1-cos(8x)]/2 and did l'hopitals rule twice (making sure I kept my fractions seperate)

9. Sep 7, 2006

### Hurkyl

Staff Emeritus
You're all making these way too hard. Think about it -- what is the intuitive meaning of:

$$\lim_{x \rightarrow 0} \frac{\sin x}{x}$$

? Does it tell you anything interesting geometrically or algebraically?

Highlight if you need to see the answer: (it tells you that, when x is small, sin x looks and acts very much like x)

So what does that tell you about these limits?

Last edited: Sep 7, 2006
10. Sep 7, 2006

### HallsofIvy

The simplest method would be L'Hopital's rule. Have you tried that? If you don't want to do that, you might separate it as
$$4\frac{sin(4x)}{4x}\frac{sin(4x)}{tan(x)}$$
You should know the limit of $4\frac{sin(4x)}{4x}$ and you should be able to use the trig identity sin(4x)= 2sin(2x)cos(2x) to reduce
$$\frac{sin(4x)}{tan(x)}$$

11. Sep 7, 2006

### HallsofIvy

This same question was asked under "homework". I am going to combine the two threads.

Last edited by a moderator: Sep 7, 2006
12. Sep 7, 2006

### Hurkyl

Staff Emeritus
You shouldn't need the trig identity. The only facts you need are the elementary limit properties, and that the limits of (sin x)/x and (tan x)/x are both 1, as x goes to zero.