## Finding the limit of trig functions

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...
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 Recognitions: Homework Help 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?
 $$\lim_{x->\pi}\frac{\sqrt{x}}{\csc{x}}$$ $$\lim_{x->\pi}\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}}}$$ simplify and what is ... $$\frac{\sin{x}}{x}$$

Recognitions:
Homework Help

## Finding the limit of trig functions

 Quote by rocophysics $$\lim_{x->\pi}\frac{\sqrt{x}}{\csc{x}}$$ $$\lim_{x->\pi}\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}}}$$ simplify
and obtain $$\lim_{x->\pi}\sqrt{x}}\sin{x}$$,

which should be decently-behaved...

 and what is ... $$\frac{\sin{x}}{x}$$
Is this a separate question?
 if you divide the limit question by x, you obtain sinx\x which allows you to have an answer other than 0

Recognitions:
Homework Help
 Quote by rocophysics if you divide the limit question by x, you obtain sinx\x which allows you to have an answer other than 0
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??
 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...
 that's exactly what i get eventually, idk what to tell you :(
 Well, my book says the correct answer is 0 but I don't know how they arrived at that.
 Recognitions: Homework Help 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) ...

 Quote by lLovePhysics Well, my book says the correct answer is 0 but I don't know how they arrived at that.
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
 Recognitions: Homework Help 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.
 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

 Quote by Pinkbunnies52 Maybe you could set it up as lim (h-->0) of [f(x+h)-f(x)]/h the first principles
Why? That would give you a derivative, but this question does not ask for a derivative...

 Quote by Pinkbunnies52 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
No it is not. The problem is asking for an evaluation of the limit, the asnwer should be a number, not a function.
 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.
 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

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