<s>[INST] Evaluating $\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}$ I'm being asked to evaluate the limit of:
$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}$$
There aren't any solutions, so I'm just wondering if I'm going about this properly, I first factored the numerator and denominator to:
$$\frac{(x+4)(x+1)}{(x+4)(x-1)}\;.$$
then cancelled both of the $(x+4)$ and plugged in $4$ for the $x$'s to get an answer of $5/3$.
Am I doing this right?
 [/INST] $$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to 4}\frac{x^2+x+4x+4}{x^2-x+4x-4}=$$
$$=\lim_{x\to 4}\frac{x(x+1)+4(x+1}{x(x-1)+4(x-1)}=\lim_{x\to 4}\frac{(x+1)(x+4)}{(x-1)(x+4)}=\lim_{x\to 4}\frac{(x+1)}{(x-1)}=\frac{5}{3}$$
Directly
$$\lim_{x\to 4}\frac{x^2+5x+4}{x^2+3x-4}=\frac{40}{24}=\frac{5}{3}$$
and
$$\lim_{x\to -4}\frac{x^2+5x+4}{x^2+3x-4}=\lim_{x\to -4}\frac{(x+1)}{(x-1)}=\frac{3}{5}$$
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