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There may arise a case where:

${\underset{{a\rightarrow 0}}{{\text{lim}}}\frac{f(x)}{g(x)}=\frac{0}{0}}$ or ${\underset{{a\rightarrow 0}}{{\text{lim}}}\frac{f(x)}{g(x)}=\frac{\infty }{\infty }}$

In which the limit can not be defined. \ If this is the case, we can employ the power of the L’Hôpital Rule, which is:

\begin{equation*}

{\underset{{a\rightarrow 0}}{{\text{lim}}}\frac{f(x)}{g(x)}=\underset{{a\rightarrow

0}}{{\text{lim}}}\frac{{f}^{‘}(x)}{{g}^{‘}(x)}}

\end{equation*}

L’Hospital’s Rule states that if we have an indeterminate form ${\frac{0}{0}}$ or ${\frac{\infty }{\infty }}$, we can differentiate the numerator and denominator and then define the limit.

#### Example

Find the limit of:

\begin{equation*}

{\underset{{x\rightarrow 1}}{{\text{lim}}}\frac{6x^{{4}}-5x^{{2}}-1}{\text{10}-9x^{{2}}-x}}

\end{equation*}

Following standard practice of taking a limit, we get:

\begin{equation*}

{\underset{{x\rightarrow 1}}{{\text{lim}}}\frac{6(1)^{{4}}-5(1)^{{2}}-1}{\text{10}-9(1)^{{2}}-1}=\frac{0}{0}}

\end{equation*}

In this case we get an indeterminate for the limit. If we had enough time, we could factor the numerator and denominator, simplify, and solve for the limit. However, using the L’Hospital Rule is much faster.

First we need to determine ${{f}^{‘}(x)}$ and ${{g}^{‘}(x)}$:

\begin{equation*}

{{f}^{‘}(x)=\text{24}x^{{3}}-\text{10}x}

\end{equation*}

\begin{equation*}

{{g}^{‘}(x)=-\text{18}x-1}

\end{equation*}

Using the L’Hospital Rule and plugging these functions in we get:

\begin{equation*}

{\underset{{x\rightarrow

1}}{{\text{lim}}}\frac{\text{24}x^{{3}}-\text{10}x}{-\text{18}x}=\frac{\text{24}(1)^{{3}}-\text{10}(1)}{-\text{18}(1)-1}=\frac{\text{14}}{-\text{19}}}

\end{equation*}