limit as x approaches 0 of (1-cos5x)/7x^2 ?

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limit as x approaches 0 of (1-cos5x)/7x^2 ?

Mathematics
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Do 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.
No, how is the form of this 1/2? I know that 1-cosax / ax is 0, but that doesn't seem helpful.
One way or another you need to use the derivative of cos x. Do you know l'Hopital's rule?

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or the Taylor/McLauren series of cos x?
Yes, but we haven't gotten to it in class yet, so I'm trying to get there without it.
how about the taylor series of cos x?
No
Well, 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 \[ \frac{25}{14}. \]
Yes, 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.
By algebraically, I meant manipulation not involving derivatives/l'hopital's rule, that is.
Yes, 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.
(To derive that series, you need to know the derivatives of cos x.)
As ever, you are a gentlemen and a scholar.

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