I have been having a rather difficult time trying to solve this limit
$$ \lim_{x\to0}\bigg(\frac{5}{x^4}-\frac{5}{x^2}\bigg) $$
So far, I have rewritten it to this point
$$ \lim_{x\to0}\bigg(\frac{-5(x+1)(x-1)}{x^4}\bigg) $$
and it's still not in an indeterminate form. I know how to use L'Hopital's Rule once it's in the correct form, I just can't seem to turn this one into a form that I can use. I keep moving on and coming back to this problem, and really it's just laughing at me at this point and I might just need a different pair of eyes to take a look.
Thank you in advance for any help!
This is much simpler, and not a case for l'Hospital. If $|x|<1/2$ (the $1/2$ is taken out of nowhere, but it should be less than one), then $$ 5(1-x^2)>5(1-(1/2)^2)=\frac{15}{4}. $$ Hence, if $|x|<1/2$, then $$ \frac{5}{x^4}-\frac{5}{x^2}=\frac{5(1-x^2)}{x^4}>\frac{15}{4x^4}. $$ Since $x\to 0$ in this problem, we can assume that $|x|<1/2$. Now, it is clear(to me, but is it to you?) that $$ \lim_{x\to 0}\frac{15}{4x^4}=+\infty. $$ It follows by comparison that the limit you look for is $+\infty$.