Basic Calculus Help: Solving a Limit Problem with Algebra

[naicidrac]

Hello everybody. I just need a little help with some very basic Calculus. Actually I need help with the Algebra part, but it is a Calculus problem. Here is the problem.
Lim x->-2 of (x^2+3x+2)/(2-|x|)

That is it and I know the answer is -1, but I cannot get that |x| out of the denomenator, I have tried to multiply by the conjugate, but that did not seem to work. Thanks in andave for any advice.

 

[wisredz]

Is it x^2+3x+2 or x^2+3x-2?

 

[Vegeta]

He must mean
\lim_{x\rightarrow -2}\frac{x^2+3x+2}{2-|x|}
Otherwise it's undefined.

 

[naicidrac]

I edited my post and Vegeta was correct.

 

[AKG]

As x approaches -2, x is negative.

 

[whozum]

Think of the absolute value funciton as a piecewise function, defined separately for positive numbers and negative numbers.

 

[wisredz]

Yes and keep in my to factor x^2+3x+2

 

[naicidrac]

I have thought of it as a piecewise function, but my instructions are to solve this algebraically. I have also factored it ((x+1)(x+2))/(2-|x|). Now what? Thanks for the help.

 

[whozum]

Defining piecewise functions is an algebraic method.

 

[wisredz]

As you approach -2 from left and right, lxl is defined as -x. Try to use this.

 

[keeti]

x is having the negative value in this case then IxI will have I-2I=-2
so the 2-IXI term will be 4 simple

 

[HallsofIvy]

No, for x close to -2, the denominator will be 2-|x|= 2-(-x)= 2+x. That's what you need. Now, what is the limit?

 

[naicidrac]

Yes, thank you guys for all the help. Now I see that as you are close to -2, |x| is defined as -x so the problem looks something like this.

lim ((x+1)(x+2))/(2-|x|)
x->-2

since we said |x| is -x we get 2-(-x) which cancels with the numerator and we are left with (x+1) and after pluging in the limit we get the answer which is -1.

Thanks for all the help.