Question

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

Answer 1

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.

Answer 2

No, how is the form of this 1/2? I know that 1-cosax / ax is 0, but that doesn't seem helpful.

Answer 3

One way or another you need to use the derivative of cos x. Do you know l'Hopital's rule?

Answer 4

or the Taylor/McLauren series of cos x?

Answer 5

Yes, but we haven't gotten to it in class yet, so I'm trying to get there without it.

Answer 6

how about the taylor series of cos x?

Answer 7

No

Answer 8

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 \[ \frac{25}{14}. \]

Answer 9

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.

Answer 10

By algebraically, I meant manipulation not involving derivatives/l'hopital's rule, that is.

Answer 11

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.

Answer 12

(To derive that series, you need to know the derivatives of cos x.)

Answer 13

As ever, you are a gentlemen and a scholar.