# Computing limits - logs vs powers

## Homework Statement

$$\frac{log (x^{2} + e^{2x})}{x + 3}$$ find the limit as x_> infinty

powers beat logs

## The Attempt at a Solution

going by the powers beat logs idea - simply, the limit as n-> infinity is 0.

is this correct? can you simply say that powers beat logs always?

No it's not that simple. Intuitively e^2x is going to dominate the bracket in the numerator and log(e^2x) can be easily simplified. Can you sandwich x^2+e^2x between multiples of e^2x (at least for large x)? Try the sandwich theorem.

ok - it will go to 2 then?

$$\frac{log (e^{2x})}{x + 3} \leq \frac{log (x^{2} + e^{2x})}{x + 3} \leq ?$$

$$\frac{log (e^{2x})}{x + 3} = \frac{2x}{x + 3}$$ which tends to 2 as x tends to infinity

is this correct - what should i use for thew upper bound?

Last edited:
Fine so far. How would you compare x^2 and e^2x? (at least for large x).

e^2x is considerably larger than x^2 ?

if x^2 divereges - so must e^2x?

cant say im 100% sure what your ginting towards.

Sorry if I'm being too vague.

x^2<e^2x so x^2+e^2x<2e^2x which gives you another useful bound :)

brilliant! - thanks

http://img223.imageshack.us/img223/2827/167uqnbqe2.gif [Broken]

Last edited by a moderator: