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!

  • $\begingroup$ The expression whose limit is taken is $$\frac{5(1 - x^2)}{x^4}.$$ Now, apply l'Hopital as suggested (well, twice). $\endgroup$ – Travis Willse Sep 12 '15 at 5:14
  • $\begingroup$ Is it allowed to use l'Hospital on limits that are not on the form $0/0$ (or $+\infty/+\infty$ or...)? Here you have $\text{something finite}/0$... $\endgroup$ – mickep Sep 12 '15 at 5:29
  • $\begingroup$ The limit tends to infinity ... L'Hospital's Rule does not really apply here. $\endgroup$ – Mark Viola Sep 12 '15 at 5:30
  • $\begingroup$ It blows up. Let $x=1/1000$. $\endgroup$ – André Nicolas Sep 12 '15 at 5:31
  • 1
    $\begingroup$ @mickep Although irrelevant for this problem, L'Hospital's Rule does apply to $L/\infty$ where $L$ need not be $\infty$. See NOTE at the end of this section. $\endgroup$ – Mark Viola Sep 12 '15 at 5:33

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 $+\infty$.

  • $\begingroup$ This does make some sense to me and I can follow the logic to a degree, but why do you choose a value less than 1? I know we're trying to approach 0, so I can see why we want small numbers and I can also see why we don't want to choose 1, but I'm hazy on why it would be some number 0 < x < 1? $\endgroup$ – Frank A. Sep 12 '15 at 5:37
  • $\begingroup$ It is a common idea that when you work with a limit $x\to 0$ (or some other value), it suffices to consider $x$ close to $0$ (or the other value). In this case, $1-x^2<0$ if $|x|>1$, and that would make my argument invalid. By considering only $|x|<1/2$, we can bound the function from below as was done. $\endgroup$ – mickep Sep 12 '15 at 5:40
  • $\begingroup$ Ohh okay, I see what you mean now. This was in our L'Hospital's Rule section of homework and I'm seeing my professor threw a curve ball in (being as this one doesn't use the rule to solve it). No wonder I was going in circles :) Thank you, I really appreciate your help! $\endgroup$ – Frank A. Sep 12 '15 at 5:43

This limit doesn't exist finitely :PROOF=>
Please note that $\lim_{x\to0}\bigg(\frac{5}{x^4}-\frac{5}{x^2}\bigg)$=$\lim_{x\to0}\bigg(\frac{-5(x+1)(x-1)}{x^4}\bigg)$
Now suppose(if possible) that the limit exists finitely .
${-5(x+1)(x-1)}$=$\bigg(\frac{-5(x+1)(x-1)}{x^4}\bigg)$*$x^4$[since x is not zero]
Now taking lim as x tends to zero on both sides,LHS becomes 5.How about RHS?Here you can apply Limit rules (since both the limits viz.$x^4$ and $\bigg(\frac{-5(x+1)(x-1)}{x^4}\bigg)$ exist as x tends to zero)
So RHS is zero as x tends to zero.

LHS is 5 but RHS is zero==> a contradiction.So $\lim_{x\to0}\bigg(\frac{5}{x^4}-\frac{5}{x^2}\bigg)$ doesn't exist finitely.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.