# A Limit

1. Apr 1, 2005

### Spectre5

I know that the limit as cos(t) goes to infinity is undefined becuase cosine oscillates between plus and minus one.

Now I have this limit to compute:

Limit of [ (t * cost(t)) / (e^(t)) ] as t goes to infinity

I know that the answer is 0 and I intuitively know why (because exp raises value far quicker than just t)

But, how do I go about proving this...the top is undefined and using LoHospitals rule gets no where becuase a (t * sin(t)) will still be in the numerator.

So how do I go about doing this? Can I just ignore the affects of the cos and just use LoHospitals rule for t/e^t??

Thanks for any help

2. Apr 1, 2005

### whozum

The limit of the product is the product of the limits.
Try separating your limit into things you can work with, and go from there.

3. Apr 1, 2005

### Jameson

You could use the formal Delta-Epsilon definition of a limit, but as you said earlier, you can intuitively look at the limit. Sine will oscillate between -1 and 1 forever, but x will continue to grow infinitely large. So the result will be that y-values of the function will get increasingly smaller for both positive and negative numbers. It will jump between very small positive and very small negative numbers, but both numbers are going to 0.

Last edited: Apr 1, 2005
4. Apr 1, 2005

### dextercioby

Use this

$$-\frac{t}{e^{t}}\leq \frac{t\cos t}{e^{t}}\leq +\frac{t}{e^{t}}$$

and then the "sqeeze theorem".

Daniel.

5. Apr 1, 2005

### Spectre5

thanks for the help everyone

dextercioby: ahhhh forgot about the sqeeze therom :) thanks

6. Apr 1, 2005

### dextercioby

That is "squeeze" :tongue2:

Rats!!I hate mis-spelling :yuck:

Daniel.

7. Apr 1, 2005

### Spectre5

Ya...and mine is not "sqeeze therom" but "squeeze theorem"....

I did both wrong haha...