limits

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March 2nd 2010, 12:24 PM
larryboi7

limits

lim as x approaches 0 $\frac{1}{x^4} - \frac{1}{x^2}$

I know im suppose to use l'hopitals rule but i dont see how to use it.
March 2nd 2010, 12:34 PM
Aryth

It's actually easier than you think, let's look at your problem:

$\lim_{x \to 0} \left(\frac{1}{x^4} - \frac{1}{x^2}\right)$

Recall the distribution over subtraction property to get:

$\lim_{x \to 0}\frac{1}{x^4} - \lim_{x \to 0} \frac{1}{x^2}$

Now, you can solve these one at a time with L'Hopital and then subtract them, I'll work out the second one:

$\lim_{x \to 0}\frac{1}{x^2}$

Apply L'Hopital's rule:

$\lim_{x \to 0} \frac{0}{2x} = 0$

I think you can do the other one.
March 2nd 2010, 12:50 PM
General

Quote:

Originally Posted by Aryth

It's actually easier than you think, let's look at your problem:

$\lim_{x \to 0} \left(\frac{1}{x^4} - \frac{1}{x^2}\right)$

Recall the distribution over subtraction property to get:

$\lim_{x \to 0}\frac{1}{x^4} - \lim_{x \to 0} \frac{1}{x^2}$

Now, you can solve these one at a time with L'Hopital and then subtract them, I'll work out the second one:

$\lim_{x \to 0}\frac{1}{x^2}$

Apply L'Hopital's rule:

$\lim_{x \to 0} \frac{0}{2x} = 0$

I think you can do the other one.

You can not do that.
$\lim_{x\to 0} \frac{1}{x^2}$ is not one of the L`Hospital`s Rule cases.
Also, if $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ are infinite, then (in general):
$\lim_{x\to a} \left( f(x) - g(x) \right) {\color{red} \neq } \lim_{x\to a} f(x) - \lim_{x\to a} g(x)$

For OP's problem:
$\lim_{x\to 0} \left( \frac{1}{x^4} - \frac{1}{x^2} \right) = \lim_{x\to 0} \frac{1}{x^2} \left( \frac{1}{x^2} - 1 \right) = \infty ( \infty - 1 ) = \infty$
and this means the y-axis is a vertical asymptote for the graph of the function $f(x)=\frac{1}{x^4} - \frac{1}{x^2}$ .