Finding Limits: ln and L'hopital's Rule Explained

[ada0713]

Homework Statement

lim[p\rightarrow0] \frac{ ln(2x^{p}+3y^{p})}{p^2}

Homework Statement

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)

 

[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

 

[Office_Shredder]

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)