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$$\lim _{x\to 2}\:\frac{x}{x^2-4}$$

It diverges, by differentiating you get 1/4, but the answer isn't 1/4

In what conditions i can't use L'hopital rule?

Thanks...

problematic
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    It's not a limit of the indeterminate form $0/0$. – J.R. Oct 24 '14 at 16:10
  • It's only applicable in cases of indeterminates. $2/0$ is not indeterminate, since that diverges no matter what. $0/0$ is, however, because sometimes it doesn't diverge (such as $x/x$ when $x$ goes to $0$). – Akiva Weinberger Oct 24 '14 at 19:57

3 Answers3

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You can't use L'Hospital because you don't have an indeterminate form $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$. Notice that if $x \to 2⁺$, then $x^2 - 4 > 0$, and this way $\lim _{x\to 2^+}\:\frac{x}{x^2-4} = +\infty$. On the other hand, $x \to 2^-$ means that $x < 2$, then $x^2 - 4 < 0$, and so $\lim _{x\to 2^-}\:\frac{x}{x^2-4} = -\infty$. Since $$\lim _{x\to 2^+}\:\frac{x}{x^2-4} \neq \lim _{x\to 2^-}\:\frac{x}{x^2-4},$$ the limit doesn't exist.

Ivo Terek
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The given fraction is not of any indeterminate form since the numerator is equal to $2$ as $x \to 2$, so the use of L' Hopital's rule is not justified.

Jimmy R.
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L'Hopital's rule only applies if when you evaluate the numerator of the expression at the limit, $2$ in this case, you get that they are either both zero, or both infinite. Since when you evaluate the numerator at $2$ you get $2$, you cannot use L'Hopital's rule.

Mike Pierce
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