Obnoxious Posted April 21, 2005 Share Posted April 21, 2005 Okay, can someone help me with this little puppy? [math]\lim_{x\to0}(\frac{1}{sin x}-\frac{1}{x})[/math] Link to comment Share on other sites More sharing options...
Dapthar Posted April 21, 2005 Share Posted April 21, 2005 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. Link to comment Share on other sites More sharing options...
Obnoxious Posted April 22, 2005 Author Share Posted April 22, 2005 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. Link to comment Share on other sites More sharing options...
Dapthar Posted April 22, 2005 Share Posted April 22, 2005 I don't wanna post all my steps, but is the answer 0? Yup. Link to comment Share on other sites More sharing options...
Dave Posted April 22, 2005 Share Posted April 22, 2005 Just a quick hint: To get [math]\sin(x)[/math] instead of [math]sin(x)[/math], use \sin - looks a lot nicer Link to comment Share on other sites More sharing options...
Obnoxious Posted April 23, 2005 Author Share Posted April 23, 2005 Hooray!! I r winnar! Link to comment Share on other sites More sharing options...
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