You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Limit, x->0, [(cxc(x))^2 - 1/x^2]^2 w/o Taylor series
- Thread starter Xilef07
- Start date
Re: Limit help please.
Is that \(\displaystyle \frac{csc^{2}(x)-1}{x^{2}}\) or \(\displaystyle csc^{2}(x)-\frac{1}{x^{2}}\)
Please use proper grouping symbols so there is no ambiguity.
The first one diverges and is easier. The second converges, but is a little tougher.
Is that \(\displaystyle \frac{csc^{2}(x)-1}{x^{2}}\) or \(\displaystyle csc^{2}(x)-\frac{1}{x^{2}}\)
Please use proper grouping symbols so there is no ambiguity.
The first one diverges and is easier. The second converges, but is a little tougher.
Re: Limit help please.
Yes, I know. I was afraid it was that one. Are you familiar with L'Hopital's rule?.
Let's work on the inside without the square.
Rewrite \(\displaystyle csc^{2}(x)-\frac{1}{x^{2}}=\frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}\)
Now, if you know L'Hopital we can continue. It may not be a good idea if you're not and we may have to find another method.
Yes, I know. I was afraid it was that one. Are you familiar with L'Hopital's rule?.
Let's work on the inside without the square.
Rewrite \(\displaystyle csc^{2}(x)-\frac{1}{x^{2}}=\frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}\)
Now, if you know L'Hopital we can continue. It may not be a good idea if you're not and we may have to find another method.
Re: Limit help please.
Yes, you can do that with L'Hopital. Sometimes you have to.
Since we rewrote as \(\displaystyle \frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}\)
Now, derivatives of top and bottom.
1st time: \(\displaystyle \frac{2x-2sin(x)cos(x)}{2x^{2}sin(x)cos(x)+2xsin^{2}(x)}\)
2nd time: \(\displaystyle \frac{4-4cos^{2}(x)}{(4x^{2}-2)cos^{2}(x)+8xsin(x)cos(x)-2x^{2}+2}\)
3rd time: \(\displaystyle \frac{8sin(x)cos(x)}{(12x^{2}-8x^{2})sin(x)cos(x)-24xsin^{2}(x)+12x}\)
4th time: \(\displaystyle \frac{16cos^{2}(x)-8}{(48-16x^{2})cos^{2}(x)-64xsin(x)cos(x)+8x^{2}-24}\)
Now, we could have simplified as we went, but I left it spread out so it's easier to see. You are ready to take your limit. Can you see it now?. Don't worry about the square. You can just square your result.
With some ingenuity, we may be able to do it without L'Hopital, but this works OK.
Yes, you can do that with L'Hopital. Sometimes you have to.
Since we rewrote as \(\displaystyle \frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}\)
Now, derivatives of top and bottom.
1st time: \(\displaystyle \frac{2x-2sin(x)cos(x)}{2x^{2}sin(x)cos(x)+2xsin^{2}(x)}\)
2nd time: \(\displaystyle \frac{4-4cos^{2}(x)}{(4x^{2}-2)cos^{2}(x)+8xsin(x)cos(x)-2x^{2}+2}\)
3rd time: \(\displaystyle \frac{8sin(x)cos(x)}{(12x^{2}-8x^{2})sin(x)cos(x)-24xsin^{2}(x)+12x}\)
4th time: \(\displaystyle \frac{16cos^{2}(x)-8}{(48-16x^{2})cos^{2}(x)-64xsin(x)cos(x)+8x^{2}-24}\)
Now, we could have simplified as we went, but I left it spread out so it's easier to see. You are ready to take your limit. Can you see it now?. Don't worry about the square. You can just square your result.
With some ingenuity, we may be able to do it without L'Hopital, but this works OK.
Re: Limit help please.
Is there a reason why you wrote it the way you wrote? Because I noticed that you expanded sin^2 or cos^2 to (1-cos^2) or (1-sin^2). Also, if I was to just differentiate it using the product rule the result would be much longer. So how did you know to write it the way that you did? I'm sorry if that's an unclear question.
Is there a reason why you wrote it the way you wrote? Because I noticed that you expanded sin^2 or cos^2 to (1-cos^2) or (1-sin^2). Also, if I was to just differentiate it using the product rule the result would be much longer. So how did you know to write it the way that you did? I'm sorry if that's an unclear question.
Re: Limit help please.
With all the trig identities out there, there are many ways to write it. I just wrote it that way hoping it would allow you to see it better. If you merely plug in x=0 in the last one you should get your limit. Then don't forget to square it.
Maybe you could try something without L'Hopital. May be tricky though. Note the difference of two squares in the numerator.
\(\displaystyle \frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}=\frac{(x+sin(x))(x-sin(x)}{x^{2}sin(x)}\)
With all the trig identities out there, there are many ways to write it. I just wrote it that way hoping it would allow you to see it better. If you merely plug in x=0 in the last one you should get your limit. Then don't forget to square it.
Maybe you could try something without L'Hopital. May be tricky though. Note the difference of two squares in the numerator.
\(\displaystyle \frac{x^{2}-sin^{2}(x)}{x^{2}sin^{2}(x)}=\frac{(x+sin(x))(x-sin(x)}{x^{2}sin(x)}\)