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Consider the first example using repeated l'Hôpital:

$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = ... = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}(24x)}{\frac{d}{dx}(24x)} = \frac{24}{24}=1 $$

Consider the following example using a different method:

$$ \lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0}\frac{\frac{x^4}{x^4}}{\frac{x^4}{x^4}+\frac{x^2}{x^4}} = \lim_{x \rightarrow 0} \frac {1}{1 +\frac{1}{x^2}} = \frac {1}{1+\infty} = \frac{1}{\infty}=0 $$

The graph here clearly tells me the limit should be $0$, but why does l'Hôpital fail?

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  • 7
    $\begingroup$ These $...$ in the first line is confusing. How many times are you performing L'Hospital Rule in the first case? $\endgroup$ – imranfat Nov 1 '17 at 1:03
  • $\begingroup$ it is a small L before the ' $\endgroup$ – mathreadler Nov 1 '17 at 4:28
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    $\begingroup$ You also forgot to apply the simplest method: $\displaystyle\lim_{x\to0}\frac{x^4}{x^4+x^2}=\lim_{x\to0}\frac{x^2}{x^2+1}=0$ (where you can't apply l’Hôpital). $\endgroup$ – egreg Nov 1 '17 at 10:59
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    $\begingroup$ When you write "..." without being precise about what it means (at least in your own head), you ask for trouble. $\endgroup$ – user21820 Nov 1 '17 at 16:38
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    $\begingroup$ This might be worth reading: What to check when using L'Hospital $\endgroup$ – Hirshy Nov 3 '17 at 13:40
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$$\lim_{x \rightarrow 0} \frac{x^4}{x^4+x^2} = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}(x^4)}{\frac{d}{dx}(x^4+x^2)} = \lim_{x \rightarrow 0} \frac{4x^3}{4x^3+2x} = \lim_{x\to0} \frac{12x^2}{12x^2+2} = \frac{0}{0+2} = 0$$

There. You can't apply l'Hospital there because the denominator doesn't go to $0$.

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You haven't checked whether L'Hopital could be applied each time.

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After doing derivative one more time you get $12x^2 +2 $ which is not $0$ when $x$ goes to $0$.

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