Answer
L'Hospital's rule does not apply.
limit = $3$
Work Step by Step
Theorem 11.3 L'Hospital's Rule:
If $f$ and $g$ are differentiable functions such that
substituting $x=a$ in the expression $\displaystyle \frac{f(x)}{g(x)}$ gives
the indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then $\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \displaystyle \frac{f^{\prime}(x)}{\mathrm{g}^{\prime}(x)}$.
That is, we can replace $f(x)$ and $g(x)$ with their derivatives and try again to take the limit.
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When x approaches $1$, (evaluating for x=$1$)
the numerator approaches $6$,
the denominator approaches $2$,
$\displaystyle \frac{f(x)}{g(x)}$ is not of indeterminate form.
L'Hospital's rule does not apply
By evaluation, the limit equals $\displaystyle \frac{6}{2}=3$
