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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?

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can you please share your method? –  lab bhattacharjee Feb 16 '13 at 12:56
    
It was my mistake, I've decomposed both functions in the wrong way. When I do it like $(3x^2+5x-2)=3(x+2)(x-\frac{1}{3})$, then it's fine. –  REACHUS Feb 16 '13 at 12:59
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2 Answers 2

up vote 6 down vote accepted

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

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

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

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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 $$\frac{12-10-2}{16-18+2}$$

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should not we point out the form $\frac00$ in the 1st expression? –  lab bhattacharjee Feb 16 '13 at 13:01
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