Calculating simple limit: $\lim_{x \to -2} \frac{(3x^2+5x-2)}{(4x^2+9x+2)}$

Question

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?

Answer 1

$$ \frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\frac{(x+2)(3x-1)}{(x+2)(4x+1)}=\frac{(3x-1)}{(4x+1)}\text{ if }x+2\ne0$$

As $x\to-2,x\ne-2\implies x+2\ne0$

So, $$\lim_{x\to-2}\frac{(3x^2+5x-2)}{(4x^2+9x+2)}=\lim_{x\to-2}\frac{(3x-1)}{(4x+1)}=\frac{3(-2)-1}{4(-2)+1}=1$$

Answer 2

Make L'hospital and you have $$\lim_{x\rightarrow-2} \frac{3x^2+5x-2}{4x^2+9x+2}=\lim_{x\rightarrow -2}\frac{6x+5}{8x+9}=\frac{-12+5}{-16+9}=1$$ Edit at first you have to show you have a $\frac{0}{0}$ expression. For $x=-2$ we have $$\frac{12-10-2}{16-18+2}$$