Solving Limit Question: x^(-2) - (cosecx)^2

[wimma]

Homework Statement

lim as x tends to zero of (x^(-2) - (cosecx)^2)

Homework Equations

L'Hospital's rule

The Attempt at a Solution

I keep getting things in the denominator that won't go away no matter how much I differentiate. I'm missing something here...

 

[Mark44]

Show us what you did.

 

[VietDao29]

Homework Statement

lim as x tends to zero of (x^(-2) - (cosecx)^2)

Homework Equations

L'Hospital's rule

The Attempt at a Solution

I keep getting things in the denominator that won't go away no matter how much I differentiate. I'm missing something here...

You mean:

\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right), right?

If you really want to do it, by applying L'Hospital rule, then the first step is to group them into 1 fraction, like this:

\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}

Now, just apply L'Hospital Rule, if you want us to check your work, you can always post it here, and we'll be more than willing to help you out. :)

Btw, this is actually the hard way, you have to apply the rule four-folds to get to the final result.

Have you covered Talor's expansion yet? This could make the problem way simpler.

 

[wimma]

yeah we've learned taylor polynomials. i guess i could make a taylor polynomial of appropriate degree for each one, though the question says to use l'hospitals.

i went

<br /> \lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)<br /> = \lim_{x \rightarrow 0} \left( \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{sin x cos x - x}{x ^ 2 sin x cos x + x sin ^2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{-sin^2 x + cos ^2 x -1}{x ^ 2 cos (2x) + 3 x sin (2x) + sin ^2 x} \right)<br />

then i gave up

 

[snipez90]

The denominator of the last expression on the right doesn't look correct. You should have a lot more terms.

 

[VietDao29]

yeah we've learned taylor polynomials. i guess i could make a taylor polynomial of appropriate degree for each one, though the question says to use l'hospitals.

i went

<br /> \lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)<br /> = \lim_{x \rightarrow 0} \left( \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{sin x cos x - x}{x ^ 2 sin x cos x + x sin ^2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{-sin^2 x + cos ^2 x -1}{2 x sin x cos x} \right)<br />

then i gave up

In the first post I told you to use it 4 times. You only used it twice. Since it's still in the form 0/0, just continue until you cannot go any further.

 

[wimma]

The denominator of the last expression on the right doesn't look correct. You should have a lot more terms.

yeah cos i gave up before i finished the denominator

fix'd

 

[snipez90]

Why? If you don't want to use the product rule extended to products of more than two factors, then use the fact that sin(2x) = 2(sin x)(cos x) or sin(x)cos(x) = sin(2x) / 2 and just keep adding like terms together after differentiating.

 

[wimma]

so i do it four times and get
<br /> \lim{x \rithgtarrow 0} \left( \frac{-4 cos (2x)}{-20 sin (2x) + 16 cos (2x) - 4 x ^ 2 cos (2x)} \right)<br />

 

[VietDao29]

yeah we've learned taylor polynomials. i guess i could make a taylor polynomial of appropriate degree for each one, though the question says to use l'hospitals.

i went

<br /> \lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)<br /> = \lim_{x \rightarrow 0} \left( \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{sin x cos x - x}{x ^ 2 sin x cos x + x sin ^2 x} \right)<br /> = \lim{x \rightarrow 0} \left( \frac{-sin^2 x + cos ^2 x -1}{x ^ 2 cos (2x) + {\color{red}3} x sin (2x) + sin ^2 x} \right)<br />

then i gave up

You can continue using it for 2 more times, or by looking at it a little bit closely, your final expression is:

\lim_{x \rightarrow 0} \left( \frac{-\sin^2 x + \cos ^2 x -1}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right) = \lim_{x \rightarrow 0} \left( \frac{-2\sin^2 x}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right)

Now, what will you get if you divide both numerator, and denominator by x²?

--------------

Oh, and btw, you have miscalculated it, the factor 3 in your step should read 2 instead. I have corrected it for you. :)

 

[VietDao29]

so i do it four times and get
<br /> \lim{x \rithgtarrow 0} \left( \frac{-4 cos (2x)}{-20 sin (2x) + 16 cos (2x) - 4 x ^ 2 cos (2x)} \right)<br />

Ohhhh.. :(

Since you have differentiated it incorrectly in the denominator, in your second step. I'm sorry to inform you that this is wrong. :( You have to re-calculate everything from your second use of L'Hospital's rule. :(

Or, you can try my hint in the post above. :)

 

[snipez90]

so i do it four times and get
<br /> \lim{x \rithgtarrow 0} \left( \frac{-4 cos (2x)}{-20 sin (2x) + 16 cos (2x) - 4 x ^ 2 cos (2x)} \right)<br />

I don't think this is correct but as others have noted, it's hard to help when you don't show the intermediate steps.

 

[wimma]

cheers =] i got the result

i'm looking at trying to do the same problem with taylor polynomials... i get the taylor polynomial for sin ^2 (x) to be\sum{\frac{-(-1)^k 2^{2k-1}x^{2k}}{(2k)!}} ...
inverting this to get the one for cosec ^2 (x) looks a bit mean.

 

[VietDao29]

cheers =] i got the result

i'm looking at trying to do the same problem with taylor polynomials... i get the taylor polynomial for sin ^2 (x) to be\sum{\frac{-(-1)^k 2^{2k-1}x^{2k}}{(2k)!}} ...

Ummm.. How did you get that? It doesn't look quite correct to me. Could you should us the step, so that i can help you point out where it went wrong?

inverting this to get the one for cosec ^2 (x) looks a bit mean.

No, you don't have to find the expansion for csc²(x). Just truncate it (the expansion for sin²(x)) wisely, then use that truncated expression to plug in the limit. Much simpler than to find its reciprocal, right? :)

\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}