Limits from the right and left

Question

Evaluate the limits below,

$$\lim_{x\to2^+}\frac{x-2}{x^2-4} $$ and $$\lim_{x\to2^-}\frac{x-2}{x^2-4} $$

Alright, I know that the limit from the right will equal positive infinity and the left will equal the negative infinity, by graphing.

Now, how do I solve this problem without graphing??

1)How would I solve if it approaches from both direction? Do I substitute the value?

2)How would I solve from the right/left side? How would I know without graphing?

I want to use this as an example to all my related questions. How would I solve for both side, and right/left side without graphing? Will there be any certain value showing later? Are there any useful theorems?

After factorized into,

$$ \frac{1}{x+2}$$

How is answer going to be different when approaching from right and left? or both sides?

I think you misinterpreted the graph you mentioned. The +/- infinity limits you described occur at $x=-2$, not $x=2$. — Jason C, Dec 01 '14 at 04:15
I think you should rather GRAPH the function before you declare you know the limits and ask how to calculate them without graphing... :) — CiaPan, Dec 01 '14 at 06:34

score 4 · Accepted Answer · answered Dec 01 '14 at 00:03

4

Note that the limit exists

$$\lim_{x\to2}\frac{x-2}{x^2-4}=\lim_{x\to2}\frac{x-2}{(x-2)(x+2)}=\lim_{x\to2}\frac{1}{x+2}=\frac14.$$

answered Dec 01 '14 at 00:03

mfl

29,399

score 1 · Answer 2 · answered Dec 01 '14 at 00:03

1

Hint; $\displaystyle a^2-b^2=(a+b)(a-b)$

answered Dec 01 '14 at 00:03

UserX

4,930

score 1 · Answer 3 · answered Dec 01 '14 at 00:03

1

HiNt:

$$\frac{x-2}{x^2-4} = \frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$$

answered Dec 01 '14 at 00:03

Aaron Maroja

17,571

Limits from the right and left

3 Answers3