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anonymous
 4 years ago
limit as x approaches 0 of (1cos5x)/7x^2 ?
anonymous
 4 years ago
limit as x approaches 0 of (1cos5x)/7x^2 ?

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Do you know this result? \[ \lim_{x \rightarrow 0} \frac{1  \cos x}{x^2} = \frac{1}{2} \] If so, put your equation in that form and you find your answer. Hint: the answer to your problem is not 1/2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, how is the form of this 1/2? I know that 1cosax / ax is 0, but that doesn't seem helpful.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1One way or another you need to use the derivative of cos x. Do you know l'Hopital's rule?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1or the Taylor/McLauren series of cos x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, but we haven't gotten to it in class yet, so I'm trying to get there without it.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1how about the taylor series of cos x?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Well, I'm fairly confident you need some result using the derivative of cos x. For the moment, if you accept the result I gave above, then \[ \frac{1  \cos(5x)}{7x^2} = \frac{1  \cos(5x)}{(7/25)(5x)^2} \] \[ = \frac{25}{7} \frac{1\cos(5x)}{(5x)^2} \] Now the limit of that expression on the right is 1/2, hence the limit of the entire expression is \[ \frac{25}{14}. \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, I came to the same conclusion using derivatives (and I thank you), but we haven't gotten there in class yet and I hesitate to use them. I had hoped to get there algebraically but I see no option.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By algebraically, I meant manipulation not involving derivatives/l'hopital's rule, that is.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, I understand. As I say, I don't think you can escape using the derivative one way or another. The other way besides l'Hopital's rule is the use the power series expansion, the Taylor series: \[ \cos x = 1  \frac{x^2}{2!} + \frac{x^4}{4!}  \frac{x^6}{6!} + ... \] You can see from that how the result I wrote down above could be derived.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1(To derive that series, you need to know the derivatives of cos x.)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As ever, you are a gentlemen and a scholar.
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