Thread: limit question

Why is lim(x->0) 1/(x^n), where n is even = infinity and where it's odd undefined?

For example,

lim(x->0) 1/x^2 = 0 ... why?

1/0 = indeterminant form... Derivative of 1 = 0...derivative of x^2 = 2x...

0/0 ... 0/2... = 0. Where does infinity come into play?

Originally Posted by Ideasman

Why is lim(x->0) 1/(x^n), where n is even = infinity and where it's odd undefined?

Both limits do not exist.
When it is even then both the limits from the right (positives) and limits from the left (negatives) are the same because you raise it to an even power. Thus, they approach positive infinite.
However, when it is odd the limits from the right remain positive and approach infinite. While the limits from the left remain negative and approach negative infinite.

lim(x->0) 1/x^2 = 0 ... why?

That is wrong.

I meant infinity. Why can't you use L'Hopitals.. I know what it looks like graphically. I want to know why this doesn't make sense numerically.

You have an indeterminant form, which then says you can use L'Hopitals. Why is it infinity.

Originally Posted by Ideasman

I meant infinity. Why can't you use L'Hopitals.. I know what it looks like graphically. I want to know why this doesn't make sense numerically.

You have an indeterminant form, which then says you can use L'Hopitals. Why is it infinity.

L'Hopitals rule will only give a finite limit when the thing has a finite limit.
These things do not have finite limits, there is no way to get a finite limit
out of it.

Try experimenting with the thing numericaly to see what is happening.

You will find that for both even and odd values 1/x^n blows up when
n gets small. n-even gives large positive values for positive and negative
x. n-odd gives large positive values for positive x and "large" negative values
for negative x.

(also 1/x^n is not considered an indeterminate form for the purposes of
L'hopitals rule).

RonL