# Finding the limit of trig functions

1. Sep 18, 2007

### lLovePhysics

1. $$\lim_{x \rightarrow \pi}(\frac{\sqrt{x}}{csc x})$$

From this function I know that $$csc x= \frac{1}{sin x}$$ which cannot equal 0.

X, therefore, cannot equal $$\pi n$$ where n is any integer.

Am I doing something wrong here? When I graph the function, there don't seem to be any asymptotes at all. I don't get this. I thought that the limit at pi was suppose to be equal to negative infinity or positive infinity but it isn't...

Last edited: Sep 18, 2007
2. Sep 18, 2007

### dynamicsolo

Given that $$csc x= \frac{1}{sin x}$$,

how could you re-write this expression: $$(\frac{\sqrt{x}}{csc x})$$ ?

Is there any problem in applying the Limit Laws to it?

3. Sep 18, 2007

### rocomath

$$\lim_{x->\pi}\frac{\sqrt{x}}{\csc{x}}$$

$$\lim_{x->\pi}\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}}}$$

simplify and what is ...

$$\frac{\sin{x}}{x}$$

4. Sep 18, 2007

### dynamicsolo

and obtain $$\lim_{x->\pi}\sqrt{x}}\sin{x}$$,

which should be decently-behaved...

Is this a separate question?

5. Sep 18, 2007

### rocomath

if you divide the limit question by x, you obtain sinx\x which allows you to have an answer other than 0

6. Sep 18, 2007

### dynamicsolo

What is this in reference to? I'm not following you.

The compound fraction you wrote:

$$\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}} }$$

becomes

$$\sqrt{x}}\sin{x}$$ .

Where did (sin x)/x come from??

7. Sep 18, 2007

### lLovePhysics

Okay, I got $$\lim_{x \rightarrow \pi}x\sqrt{x}$$ but then if I use direct substitution I get $$\pi\sqrt{\pi}$$ which is not the correct answer...

8. Sep 18, 2007

### rocomath

that's exactly what i get eventually, idk what to tell you :(

9. Sep 19, 2007

### lLovePhysics

Well, my book says the correct answer is 0 but I don't know how they arrived at that.

10. Sep 19, 2007

### dynamicsolo

Ah, here's the problem.

Forget what I had here before! That's what I get for trying to analyze something like this in the wee hours...

We're getting distracted by that (x/sin x). You're evaluating the limit at *x = pi*, not x = 0! Your expression becomes [sqrt(pi)]/(pi/0) = [sqrt(pi)]/inf. = 0 . So there *is* no contradiction! It just makes the algebra easier if you multiply the original function through by sin x , rather than playing with the compound fraction. But the results are equivalent.

I couldn't understand how the original function was getting turned into

x·sqrt(x) , which behaves nothing like sqrt(x)/(csc x) ...

Last edited: Sep 19, 2007
11. Sep 19, 2007

### rocomath

if it's zero, rather than putting it under x to get rid of sinx, just plug it in when you have sinx times sqrtX

12. Sep 19, 2007

### dynamicsolo

I think this illustrates the perils of too-late-night calculus. I've completely revised my previous post. There is no difficulty and the answer *is* zero.

Last edited: Sep 19, 2007
13. Oct 24, 2007

### Pinkbunnies52

Maybe you could set it up as lim (h-->0) of [f(x+h)-f(x)]/h
the first principles

or you can just know that its -cot(x)sec(x) or something like that, i dont have my limit laws sheet infront of me right now, but hopefully that can help

14. Oct 24, 2007

### d_leet

Why? That would give you a derivative, but this question does not ask for a derivative...

No it is not. The problem is asking for an evaluation of the limit, the asnwer should be a number, not a function.

15. Sep 6, 2009

### Storm Butler

wait why i dont get why the answer couldnt be that x^(1/2)*sinx as x approaches pi so that sinx goes to zero by direct substitution and 0*x^(1/2)=0.

16. Sep 6, 2009

### Elucidus

Assuming we are all clear that the original limit's argument simplifies to $$\sqrt{x} \sin x$$, the substitution of pi into the argument only works if the argument is continuous at pi. Just seeing that sin(pi) = 0 is not enough. One must also check to see what the behavior of the radical factor is. Fortunately the limit is:

$$\lim_{x \rightarrow \pi} \sqrt{x} \sin x = \sqrt{\pi} \sin \pi = \sqrt{\pi} \cdot 0 = 0$$.

(Aplogies in advance for potentially violating the "don't give out solutions" forum policy, but I am assuming the solution has already been found and I mean only to comment on being careful with the logic at arriving at that solutin.)

--Elucidus