# Applying L'Hopital's rule to $\lim\limits_{x \to 0}\frac{2}{x^2}$

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

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..

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

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