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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!

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $+\infty$.

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  • $\begingroup$ 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
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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.

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