How Do You Evaluate the Limit of (2-|x|)/(2+x) as x Approaches -2?

[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?

 

[Mark44]

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?

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

 

[vanhees71]

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

 

[JessicaJ283782]

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

 

[Mark44]

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

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

 

[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?

 

[Mark44]

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.

 

[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)?

 

[Mark44]

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)?

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.

 

[Ray Vickson]

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)?

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?