Finding the limit of trig functions

lLovePhysics

I'm having a hard time find the limits of these trig functions. Please help me with it. Thanks in advance.

1. [tex]\lim_{x \rightarrow \pi}(\frac{\sqrt{x}}{csc x})[/tex]

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

X, therefore, cannot equal [tex]\pi n[/tex] 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...

dynamicsolo

Given that [tex]csc x= \frac{1}{sin x}[/tex],

how could you re-write this expression: [tex](\frac{\sqrt{x}}{csc x})[/tex] ?

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

rocomath

[tex]\lim_{x->\pi}\frac{\sqrt{x}}{\csc{x}}[/tex]

[tex]\lim_{x->\pi}\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}}}[/tex]

simplify and what is ...

[tex]\frac{\sin{x}}{x}[/tex]

dynamicsolo

and obtain [tex]\lim_{x->\pi}\sqrt{x}}\sin{x}[/tex],

which should be decently-behaved...



Is this a separate question?

rocomath

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

dynamicsolo

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

The compound fraction you wrote:

[tex]\frac{\frac{\sqrt{x}}{1}}{\frac{1}{\sin{x}} }[/tex]

becomes

[tex]\sqrt{x}}\sin{x}[/tex] .

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

lLovePhysics

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

rocomath

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

lLovePhysics

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

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) ...

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

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.

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

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.

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.

Elucidus

Assuming we are all clear that the original limit's argument simplifies to [tex]\sqrt{x} \sin x[/tex], 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:

[tex]\lim_{x \rightarrow \pi} \sqrt{x} \sin x = \sqrt{\pi} \sin \pi = \sqrt{\pi} \cdot 0 = 0[/tex].

(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