# Calculating the limits

1. Dec 7, 2007

1. The problem statement, all variables and given/known data

lim[p$$\rightarrow$$0] $$\frac{ ln(2x^{p}+3y^{p})}{p^2}$$
1. The problem statement, all variables and given/known data

I first plugged in zero to see if i can you L'hopital's rule
I got a (constant)/0 which is not an indeterminant form.

If I can't use the rule, what should be my next step to find the limit?
( I kinda guessed that the the answer's infinity but I'm not sure)

2. Dec 7, 2007

### TalonStriker

You don't need L'hopital's rule here since the numerator is always constant... The answer is more straight forward than you think...

What's lim [x->infinity] 1/(x^2) ?

edit: didn't notice that x was to the power of p...see office shredder's comment

Last edited: Dec 7, 2007
3. Dec 7, 2007

### Office_Shredder

Staff Emeritus
The numerator isn't constant, but it approaches one...
also, the limit is as p goes to 0, not infinity.

I suspect the easiest way to be rigorous is to bound ln(...) from above and below for small enough p, which gives you that constant/0 term. It's not always infinity though... you haev to be careful of the existence and sign of ln(...) depending on what values x and y take (note they have to be non-negative, or the limit definitely doesn't exist, for example)