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

1. JamesJ

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.

2. anonymous

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

3. JamesJ

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

4. JamesJ

or the Taylor/McLauren series of cos x?

5. anonymous

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

6. JamesJ

how about the taylor series of cos x?

7. anonymous

No

8. JamesJ

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 $\frac{25}{14}.$

9. anonymous

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.

10. anonymous

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

11. JamesJ

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.

12. JamesJ

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

13. anonymous

As ever, you are a gentlemen and a scholar.