#### Using the method of L' Hospital, evaluate the limit.  $\lim _{x \rightarrow 1} \frac{\left(3 x^{2}+2 x-5\right)}{\left(x^{2}-1\right)}$Option: 1 -4Option: 2 2Option: 3 4Option: 4 0

Apply the L' Hospital's Rule, when $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, etc.$ [an intermediate form] in the following way

- Differentiate $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)},$ provided that the limit $\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ exists.

- Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

$\lim _{x \rightarrow 1} \frac{\left(3 x^{2}+2 x-5\right)}{\left(x^{2}-1\right)}=\frac{0}{0}$

Here use the L' Hospital rule and also refer to the following formula.

$\frac{d}{d x} x^{n}=n x^{n-1}$

Now, derive the following:

\begin{aligned} & \lim _{x \rightarrow 1} \frac{\left(3 x^{2}+2 x-5\right)}{\left(x^{2}-1\right)} \\ & =\lim _{x \rightarrow 1} \frac{\frac{d}{d x}\left(3 x^{2}+2 x-5\right)}{\frac{d}{d x}\left(x^{2}-1\right)} \\ & =\lim _{x \rightarrow 1} \frac{(3 \times 2 x+2)}{(2 x)} \\ & =\lim _{x \rightarrow 1} \frac{(6 x+2)}{(2 x)} \\ \end{aligned}

\begin{aligned} & =\frac{(6 \times 1+2)}{(2 \times 1)} \\ & =4 \end{aligned}