# How do you find lim (xlnx)/(x^2-1) as x->1 using l'Hospital's Rule?

Dec 13, 2016

${\lim}_{x \to 1} \frac{x \ln x}{{x}^{2} - 1} = \frac{1}{2}$

#### Explanation:

Let's put:

$f \left(x\right) = \frac{x \ln x}{{x}^{2} - 1} = f \frac{x}{g} \left(x\right)$

As:

${\lim}_{x \to 1} x \ln x = 0$

and

${\lim}_{x \to 1} \left({x}^{2} - 1\right) = 0$

we can use l'Hopital's rule stating that in such case:

${\lim}_{x \to 1} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 1} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

if such limits exists.

$f ' \left(x\right) = \ln x + 1$

$g ' \left(x\right) = 2 x$

${\lim}_{x \to 1} \frac{f ' \left(x\right)}{g ' \left(x\right)} = {\lim}_{x \to 1} \frac{\ln x + 1}{2 x} = \frac{1}{2}$