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

If we apply L'Hopital rule, Then the procedure would go: $0/2x$, and then $0/2$ which is zero. What is wrong with this application L'Hopital rule, as it clearly seems wrong..

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  • $\begingroup$ Do you have the correct form to apply that rule? $\endgroup$
    – Amzoti
    Sep 23, 2013 at 3:31
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    $\begingroup$ To apply l'H's rule you must have an indetermiante form: $\;\frac00\;,\;\;or\;\;\frac\infty\infty\;$ $\endgroup$
    – DonAntonio
    Sep 23, 2013 at 3:31
  • $\begingroup$ oh. That's why it went wrong! $\endgroup$
    – L'hoptial
    Sep 23, 2013 at 3:32
  • $\begingroup$ What is your name is it really @L'hoptial? $\endgroup$
    – user845875
    Jul 24, 2021 at 13:56

2 Answers 2

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In order to use the $0/0$ case of L'Hospital's rule, we require that both the numerator and the denominator tend to $0$ at the appropriate point. The numerator does not tend to $0$.

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L'Hopital's rule states that if $$\lim_{x \to C} f(x) = 0$$ and $$\lim_{x \to C} g(x) = 0$$ then $$\lim_{x \to C} \dfrac{f(x)}{g(x)} = \lim_{x \to C} \dfrac{f'(x)}{g'(x)}$$

Another form of the rule is:

if $$\lim_{x \to C} f(x) = \infty$$ and $$\lim_{x \to C} g(x) = \infty$$ then $$\lim_{x \to C} \dfrac{f(x)}{g(x)} = \lim_{x \to C} \dfrac{f'(x)}{g'(x)}$$

In your problem $\lim_{x \to 0}2≠ 0 $ or $\infty$ so the rule doesn't apply

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  • $\begingroup$ Your statement of L'Hopital's rule is incorrect. You need some additional hypotheses. $\endgroup$ Sep 23, 2013 at 7:30

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