Evaluating Limit: Help with First Step

[crybllrd]

Homework Statement

I had an exam last night and I was stuck on this problem.
I went home where I had more time (and resources) and still can't figure it out, and I won't get my exam back (it was a final).

$\lim_{x\to0}\frac{1}{x^{2}}-\frac{1}{xsinx}$

Homework Equations

The Attempt at a Solution

I have tried combining into one fraction, splitting up into two separate limits, I can't use L'Hopital's as it is 1/inf.
What am I missing? The rest of the exam was somewhat easy, but I didn't even know where to start!
Can someone help me work though it? I think I just need help with a first step.
Thanks!

 

[clamtrox]

If you combine them into one fraction, then it's 0/0 right? Only problem is that you have to use L'Hopital 3 times to get the result.

 

[mtayab1994]

Hello crybllrd write your expression in the following form:

$$\lim_{x\rightarrow0}-\frac{x^{2}sin(x)-xsin(x)^{2}}{x^{3}sin(x)^{2}}$$

I think it should be clear from here on.

 

[CompuChip]

You can combine ito one fraction and then apply l'hospital's repeatedly.

Or you can do a series expansion.

 

[crybllrd]

Hello crybllrd write your expression in the following form:

$$\lim_{x\rightarrow0}-\frac{x^{2}sin(x)-xsin(x)^{2}}{x^{3}sin(x)^{2}}$$

I think it should be clear from here on.

How did you get the sin(x)² in the denominator? If I multiply x²by xsin(x) I get x³sin(x).

 

[mtayab1994]

If you combine them into one fraction, then it's 0/0 right? Only problem is that you have to use L'Hopital 3 times to get the result.

I used l'hopital's rule 6 times, not 3 times.

 

[HallsofIvy]

How did you get the sin(x)² in the denominator? If I multiply x²by xsin(x) I get x³sin(x).

Why would you do that? The "least common denominator" is $x^2sin(x)$. Of course, you also do not get "[itex]x^2sin^2(x)[/tex]"!<br /> <br /> $$\frac{1}{x^2}- \frac{1}{x sin(x)}= \frac{sin(x)}{x^2 sin(x)}- \frac{x}{x^2 sin(x)}$$<br /> $$= \frac{sin(x)- x^2}{x^2 sin(x)}$$<br /> <br /> Now, you can use L'Hopital's rule on that, just once. Differentiating both numerator and denomimator separately you get<br /> $$\frac{cos(x)- 2x}{2x sin(x)+ x^2cos(x)}$$.<br /> <br /> You do NOT get "1/infinity".[/itex]

 

[crybllrd]

Ah, ok, I got -1/6.
I don't remember why exactly I didn't get a good answer on the exam, but I think I was on like my 6th L'Hopital iteration and wasn't willing to spend any more time on a 4 point problem.
Thanks guys, now I can rest easy this summer, as my brain is completely fried now!

 

[sweet springs]

Hi. As already other colleagues have stated,

$\lim_{x\to0}[\frac{1}{x^{2}}-\frac{1}{xsinx}]$

$=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{x}{sinx}]$

$=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{1}{1 - \frac{x^{2}}{3!} + ...} ]$

$=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - 1 - \frac{x^{2}}{3!} + ...]$