L'Hopital's Rule Question

1. Sep 29, 2008

mk2munky

I had a test today and there was a LR limit on it that I didn't get. I thought he said he got them out of the book, but I didn't see it anywhere. The equation was:

lim x$$\rightarrow$$$$\infty$$ of (1 + cosx(1/x))x

So, I said that lim x$$\rightarrow$$$$\infty$$ of cos(1/x) = cos(0) = 1

And therefore:

lim x$$\rightarrow$$$$\infty$$ of (1 + 1)$$\infty$$ = 2$$\infty$$

But the only LR indeterminate for near that is 1$$\infty$$ type... I left it at that. I'm pretty sure my answer is incorrect. Anyone?

2. Sep 29, 2008

NoMoreExams

You wrote cosx(1/x), pretty sure you meant cos(1/x)?

Anyways, I think you are correct, take ln of both sides. You now have

$$ln|y| = x \cdot ln \left(1 + cos \left( \frac{1}{x} \right) \right) = \frac{ln \left(1 + cos \left( \frac{1}{x} \right) \right)}{\frac{1}{x}}$$

If you now evaluate the limit you get

$$\frac{ln(2)}{0}$$

Which as you pointed out is not in the form 0/0, note that if you had $$1^{\infty}$$ then you would have had ln(1)/0 = 0/0