L'Hospital's rule exception $$ \lim_{x\to-2}\frac{x^3-3x^2-10x}{(x+2)^2}$$ in this problem I tried using L'Hospital's rule but the outcome is not true why? Are there any exceptions to this rule?
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.
We find that
$$
\begin{align}
\lim_{x\to -2} \frac{ x^3 - 3x^2 - 10x }{ (x+2)^2 } &amp;= \lim_{x\to -2} \frac{ 3x^2 - 6x - 10  }{ 2(x+2)  } \\
&amp;= \pm\infty,
\end{align}
$$
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 
$$ \lim_{x \to -2} \frac{14}{x+2} = \pm\infty. $$