TITLE: solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$
QUESTION [5 upvotes]: I need to find
$$\lim_{x\rightarrow -5}  \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5}  \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5}  \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$
WAY II:
$$\lim_{x\rightarrow -5}  \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5}  \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5}  \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$
What am I doing wrong here?
Thanks!

REPLY [7 votes]: Numerator
$2x^2-50=2(x-5)(x+5)$.
Denominator
$2x^2+3x -35 =(2x-7)(x+ 5)$
$\dfrac{2(x-5)(x+5)}{(2x-7)(x+5)}=$
$\dfrac{2(x-5)}{2x+7}.$
Take the limit $x \rightarrow -5.$
Try to factorize the original expression.The term $(x+5) $ cancels out .