How do you solve this limit without L'Hospitals?

1. Jan 17, 2014

JessicaJ283782

limit as x approaches -2

(2-absolute value x)/(2+x)

How Would you check the limit from the right? I get it would be

2-x/2+x

But how would you solve it? I get the limit from the left would be 1, but how would the other side be one? We can't use L'Hospitals, and Wolframalpha says the answer is 1?

2. Jan 17, 2014

Staff: Mentor

When x is close to -2, |x| = -x.

3. Jan 17, 2014

vanhees71

You can assume $x<0$, because you take the limit $x \rightarrow -2$. Then, you do not need de L'Hospital's rule :-).

4. Jan 17, 2014

JessicaJ283782

So it would be -x for both the left and right limits? not -x for one and x for the other?

5. Jan 17, 2014

Staff: Mentor

Yes to first question. You can assume that x will be close to -2, so |x| = -x for both one-sided limits.

6. Jan 17, 2014

JessicaJ283782

Thank you! Just to make sure I understand, why is it not using (-x) for one side and (x) for the other? All of the examples she did in class said to take (-x) and (x), or one negative and positive, when checking the limits?

7. Jan 17, 2014

Staff: Mentor

Because |x| = x if x ≥ 0, and |x| = -x if x < 0. Since x is "near" -2, then it's not close to zero, so |x| will be -x on either side of -2.

8. Jan 17, 2014

JessicaJ283782

Thank you! So if the "a" value is a positive number, then you have to check both the negative and positive? (So if it was 2, you would have to check (x) and (-x)?

9. Jan 17, 2014

Staff: Mentor

No. In that case, x would be "near" 2, which means that numbers slightly smaller than 2 and numbers slightly larger than 2 would be positive.

The only time you would have to be concerned would be if the limit was as x approaches 0. Hopefully, it's clear now.

10. Jan 17, 2014

Ray Vickson

You are just confusing yourself. Start again, and this time first draw a graph of the function $f(x) = |2-x|$. When $x$ remains near -2, what is the form of the graph? Now do you see what is happening?