How can 2/sinX have a limit of infinity for X->0?

[Ken]

The following problem supposedly has a limit of infinity.

Lim as X ->0 from the right of 2/sinX

I don't see a limit that exists. I see a function that increases and decreases without bound. I also don't see a way to change the original equation.

 

[stapel]

Look at the graph of the sine wave. What value does it approach as x approaches zero (from either side)? :idea:

What is the limit, as x -> 0⁺, of 2/x? :wink:

Eliz.

 

[galactus]

\(\displaystyle \L\\\frac{2}{sin(x)}=2csc(x)\)


\(\displaystyle \L\\2\cdot\lim_{x\to\0^{+}}csc(x)=?\)


Yes, you are onto something. csc(x) has lots of 'unboundedness'.

 

[Ken]

when I graph either 2/sinX or 2cscX, I get at graph that looks like a graph of a secant as you might expect. In either case, I see a lot of verticle asymtotes that occur on the open interval. I still don't see a limit and I still don't understand how a limit at infinity can exist.

 

[stapel]

when I graph either 2/sinX or 2cscX, I get at graph that looks like a graph of a secant as you might expect.

How is the secant coming into play? (The cosecant would be understandable, but I'm not seeing how you're working the secant curve in...?)

What do you see (near and at zero) when you graph the basic sine wave?

In either case, I see a lot of verticle asymtotes that occur on the open interval.

You see a lot of vertical asymptotes on which open interval? You're looking close to zero, so -pi/2 to pi/2 should be more than enough. What are you seeing? (A link to a graph might be helpful...?)

I still don't see a limit and I still don't understand how a limit at infinity can exist.

Does your book not define what is meant by "a limit at infinity"? (It should say something like "the function grows without bound, as when approaching a vertical asymptote".)

Thank you!

Eliz.

 

[pka]

when I graph either 2/sinX or 2cscX, I get at graph that looks like a graph of a secant as you might expect. In either case, I see a lot of verticle asymtotes that occur on the open interval. I still don't see a limit and I still don't understand how a limit at infinity can exist.

It cannot have a limit at zero in most treatments of limits.
But there are some textbooks that take weird liberties with standard definitions. You should read very carefully what your text has to say about “infinite limits”.

On the other hand, it may be that the answer key is simply wrong or it may be a typo.

In any case, a mathematician will say that there is no limit at zero.

 

[Ken]

First, I want to thank you for your patience.


The secant is not coming into play. That was a poor attempt on my part to describe the graph that I get when I graph 2/sin(x) or 2csc(x).

It is not a sine wave but, as x approaches 0 from the right, the graph increases w/o bound.

My book describes properties of infinite limits by saying (for the quotient)

lim f(x)=infinity as x->c and

lim g(x)= L as X->c

Then

lim g(x)/f(x)= 0 as x->c

so, in my example; g(x) = 2

and as x approaches 0 from the right, the sin (0) tends toward 0 so the limit is infinite?

I would think it would be undefined and you could expect a verticle asymptote that increases w/o bound at 0.

Thanks again.

 

[galactus]

\(\displaystyle \L\\2\lim_{x\to\0^{+}}csc(x)=\infty\)

\(\displaystyle \L\\2\lim_{x\to\0^{-}}csc(x)=-\infty\)

As you can see, the one-sided limits are different.

Thereofre, the limit does not exist.

You can see by the graph.

 

[Ken]

one sided limits

I think I see now. Your graph is what I was getting and there is the verticle asymptote at zero; increasing without bound.

I would have posted my graph if I had known how to do that.


I want to thank you folks for your help in my endeavor to understand this stuff. I know that you are taking time out of your schedule to help people like me and I want all of you to know that it doesn't go unappreciated.

I'm gonna go and try to digest what we have covered today.