<s>[INST] solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ 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!
 [/INST] For your way $1$, check the computation of your denominator, it should give you $0$ again.
For your way $2$, check your factorization in your denominator as well.
Use L'hopital's rule:
$$\lim_{x\rightarrow -5}  \frac{2x^2-50}{2x^2+3x-35}= \lim_{x\rightarrow -5}  \frac{4x}{4x+3}=\frac{-20}{-17}=\frac{20}{17}$$
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