# L'Hopital's rule graph problem

1. Jul 15, 2004

### Moose352

Why can't I solve this?
$$\lim_{x\rightarrow 0} \frac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3}$$

Last edited: Jul 15, 2004
2. Jul 15, 2004

### Staff: Mentor

What is the problem? (I don't understand your problem statement.) Are you trying to find the limit of that expression as x goes to something?

3. Jul 15, 2004

### Moose352

Yes, I need to evaluate the limit as x->0.

4. Jul 15, 2004

### Staff: Mentor

L'Hopital's Rule

Are you familiar with L'Hopital's rule? (Look it up!) It's very useful for evaluating functions with indeterminate form (like 0/0).

5. Jul 15, 2004

### Moose352

No I'm not familiar with L'Hopital's rule. I looked it up in my book (it was listed as L'Hospital's rule :rofl: ) but I'm quite sure the problem should be able to be solved without using L'Hopital's rule.

Last edited: Jul 15, 2004
6. Jul 15, 2004

### loseyourname

Staff Emeritus
If you graph the equation, it's pretty clear that the limit is infinity.

7. Jul 15, 2004

### Moose352

No, the limit is 1/4.

8. Jul 15, 2004

### cookiemonster

Why do you say that?

cookiemonster

9. Jul 15, 2004

### Moose352

Well, because L'Hopital's Rule is like 4 chapters away! So assuming that my book hasn't screwed up, I would think they wouldn't put a problem requiring knowledge of L'Hopital's Rule without introducing it first!

10. Jul 15, 2004

### cookiemonster

Well, what are you "allowed" to use?

cookiemonster

11. Jul 15, 2004

### Wong

multiply the numerator and denominator by sqrt(1+tanx)+sqrt(1+sinx). Then you should have the factor tanx-sinx in your numerator. Now let t = tan(x/2), then tan(x) = 2t/(1-t^2), sin(x) = 2t/(1+t^2). Plug into the original expression, and note that the limit of tan(x/2)/x as x tends to 0 is 1/2. (This follows from the fact that lim sin(x)/x = 1, which can be proved without L'Hospital's rule.)

12. Jul 15, 2004

### loseyourname

Staff Emeritus
I guess my graph is wrong. Perhaps I typed the equation incorrectly. Drat.

13. Jul 16, 2004

### Moose352

Hmm, I think I may have solved it. I'm really tired, so I probably made a lot of mistakes:

$$\frac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3}$$

$$=\frac{\tan(x)-\sin(x)}{x^3(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$

$$=\frac{\frac{\sin(x)}{\cos(x)}-\sin(x)}{x^3(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$

Since the limit of sin(x)/x = 1 as x->0,

$$=\frac{\frac{1}{\cos(x)}-1}{x^2(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$

$$=\frac{1-\cos(x)}{x^2\cos(x)(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})}$$

$$=\frac{\sin(x)^2}{x^2\cos(x)(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})(1+\cos(x))}$$

$$=\frac{1}{\cos(x)(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)})(1+\cos(x))}$$

$$=\frac{1}{1(\sqrt{1+0}+\sqrt{1+0})(1+1)}=\frac{1}{4}$$

Does this work? Wong, I haven't checked out your solution yet.

Last edited: Jul 16, 2004
14. Jul 16, 2004

### HallsofIvy

Staff Emeritus
$$=\frac{\sin(x)^2}{x^2\cos(x)(\sqrt{1+\tan(x)}+\sqr t{1+\sin(x)})(1+\cos(x))}$$

$$=\frac{\sin(x)^2}{x^2\cos(x)(\sqrt{1+\tan(x)}+\sqr t{1+\sin(x)})(1+\cos(x))}$$

How do you do this step? The first has cos(x), not cos2(x).

It does reduce to $$lim_{x->0}\frac{1- cos(x)}{x^2}$$ since the rest has a limit of the rest is 1/2. I would do that either by L'Hopital's rule or by replacing cos(x) by by its McLaurin series. Either way the whole limit is 1/4.

15. Jul 16, 2004

### Moose352

Not sure what your question is Halls, but if you're asking how I got from 1-cos(x) to 1-cos(x)^2, i multiplied numerator and denominator by 1+cos(x). What

16. Jul 16, 2004

### loseyourname

Staff Emeritus
You can evaluate the limit as it is just using a table or a graph. Once I typed in the equation properly, the value of f(x) approached 0.25 as x approached 0, from both sides. Why bother doing all this? Were you asked to prove that the limit is 0.25?

17. Jul 16, 2004

### hello3719

Because THIS IS MATH, not a drawing , guessing game.

18. Jul 16, 2004

### HallsofIvy

Staff Emeritus
To Moose353: Sorry, I didn't see the 1+ cos(x) in the denominator. I see nothing wrong with your solution.

To Loseyourname: Hello3719's point is that drawing a graph and observing that it APPEARS to converge to 1/4 doesn't prove that it does (as opposed to converging to 0.2500000000000000000000000000001, say).

Last edited: Jul 16, 2004
19. Jul 16, 2004

### Moose352

No problem Halls. Thanks for the help!

20. Jul 16, 2004

### loseyourname

Staff Emeritus
I know. That's why I asked if he simply needed to get an answer or if he needed to prove an answer. If no proof is needed, a shortcut never hurts.

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