# How do you find lim_(xto2) (x^3-6x-2)/(x^3-4x) using l'Hospital's Rule or otherwise?

Jun 26, 2017

The two-sided limit doesn't exist...

graph{(x^3 - 6x - 2)/(x^3 - 4x) [-4.245, 9.8, -1.71, 5.313]}

Well, you can approach it with some manipulation.

$= {\lim}_{x \to 2} \frac{{x}^{3} - 4 x - 2 x - 2}{{x}^{3} - 4 x}$

$= {\lim}_{x \to 2} {\cancel{\frac{{x}^{3} - 4 x}{{x}^{3} - 4 x}}}^{1} - \frac{2 x + 2}{{x}^{3} - 4 x}$

$= 1 - {\lim}_{x \to 2} \frac{2 x + 2}{x \left({x}^{2} - 4\right)}$

$= 1 - 2 {\lim}_{x \to 2} \frac{x + 1}{x \left({x}^{2} - 4\right)}$

$= 1 - 2 \left[{\lim}_{x \to 2} \frac{\cancel{x}}{\cancel{x} \left({x}^{2} - 4\right)} + \frac{1}{x \left({x}^{2} - 4\right)}\right]$

$= 1 - 2 \left[{\lim}_{x \to 2} \frac{1}{\left(x + 2\right) \left(x - 2\right)} + \frac{1}{x \left(x + 2\right) \left(x - 2\right)}\right]$

We really can't use L'Hopital's rule here because it's not of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, but $\frac{1}{0}$. There's no other obvious manipulation we can do.

So, plugging in $2$, the limit is undefined. From the left, it is $\infty$, and from the right, it is $- \infty$.