# Given (x−4)/ (x^2+6x−40) how do you find the limit as x approaches 4?

Sep 22, 2016

$\frac{1}{14}$

#### Explanation:

well, we can plug in $x = 4$ to see if there is a problem in the first place

for f(x) = (x−4)/ (x^2+6x−40)

f(4) = (4−4)/ (16+24−40) = 0/0

This is $\frac{0}{0}$ indeterminate, so we are free to use L'Hôpital's Rule....

.....which tells us that
lim_(x to 4) (x−4)/ (x^2+6x−40)

$= {\lim}_{x \to 4} \frac{1}{2 x + 6}$

$= \frac{1}{8 + 6} = \frac{1}{14}$

But we might also have spotted that:

f(x) = (x−4)/ (x^2+6x−40) = color(red)( (x−4)/ ((x-4)(x + 10))) = (x-4)/(x-4) * 1/(x + 10)

And

${\lim}_{x \to 4} \frac{x - 4}{x - 4} \cdot \frac{1}{x + 10}$

...using the idea that the limit of the product is the product of the limits
$= {\lim}_{x \to 4} \frac{x - 4}{x - 4} \cdot {\lim}_{x \to 4} \frac{1}{x + 10}$

$= 1 \cdot \frac{1}{14}$

I was slightly surprised not to find a two-sided limit here, as they often come out of these kinds of question. But the fact that L'Hôpital's Rule led directly to the solution, as opposed to being one step off the two-sided solution, was the spur to factorise.