L'Hospital's rule exception

Question

$$ \lim_{x\to-2}\frac{x^3-3x^2-10x}{(x+2)^2}$$ in this problem I tried using L'Hospital's rule but the outcome is not true why? Are there any exceptions to this rule?

Answer 1

The rule says that $\lim _{x \to a} \frac {f(x)} {g(x)}=\lim _{x \to a} \frac {f'(x)} {g'(x)}$ if the original limit is in the form $\frac 0 0$ and $\lim _{x \to a} \frac {f'(x)} {g'(x)}$ exists. This last condition is important. In this case this condition is not satisfied and you cannot apply the rule.

Answer 2

We find that $$ \begin{align} \lim_{x\to -2} \frac{ x^3 - 3x^2 - 10x }{ (x+2)^2 } &= \lim_{x\to -2} \frac{ 3x^2 - 6x - 10 }{ 2(x+2) } \\ &= \pm\infty, \end{align} $$ because $$ \frac{ 3x^2 - 6x - 10 }{ 2(x+2) } = \frac{1}{2} \left( 3x - 12 + \frac{14}{x+2} \right), $$ and $$ \lim_{x \to -2} 3x-12 = -18, $$ which is finite, but $$ \lim_{x \to -2} \frac{14}{x+2} = \pm\infty. $$

Answer 3

The given function simplifies to

$$\frac{x (x-5)}{x+2} $$

The graph has two asymptotes, one is vertical at $x=-2.$

You did not have $\frac{0}{0}, \frac{\infty}{\infty} $ forms so why did you think of L'Hospital's rule? It is not an exception.