Okay, can someone help me with this little puppy?

[math]\lim_{x\to0}(\frac{1}{sin x}-\frac{1}{x})[/math]

Okay' date=' can someone help me with this little puppy?

[math']\lim_{x\to0}(\frac{1}{sin x}-\frac{1}{x})[/math]

Intuitively, as [math]x\to0[/math], [math]sin x \approx x[/math], therefore, the limit should be [math] 0 [/math].

 

At the moment, I'll just give you a hint. If you combine the two fractions, you get [math]\lim_{x\to0}{\frac{x-sin x}{x sin x}}[/math]. This is an indeterminate form (namely [math]\frac{0}{0}[/math]), so you can apply L'Hopital's rule. You'll need to apply it twice, but you'll end up getting that the limit is [math] 0 [/math].

 

If you haven't learned L'Hopital's rule yet, mention so in a post, and I'll try to compute the limit without using it.

I don't wanna post all my steps, but is the answer 0? Please tell me it's 0, as that is what I put on my test.

I don't wanna post all my steps, but is the answer 0?

Yup.

Just a quick hint: To get [math]\sin(x)[/math] instead of [math]sin(x)[/math], use \sin - looks a lot nicer

Hooray!! I r winnar!