Calculating simple limit: $\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}$ I am trying to calculate this limit:
$$
\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}
$$
What I get is $\frac{4}{3}$, however, according to Wolfram Alpha, it should be 1. What am I missing here?

$$ \frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\frac{(x+2)(3x-1)}{(x+2)(4x+1)}=\frac{(3x-1)}{(4x+1)}\text{ if }x+2\ne0$$
As $x\to-2,x\ne-2\implies x+2\ne0$
So, $$\lim_{x\to-2}\frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\lim_{x\to-2}\frac{(3x-1)}{(4x+1)}=\frac{3(-2)-1}{4(-2)+1}=1$$
