Limit when denominator = 0

Question

Can someone explain me why the following is not defined:

$$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)} = \text{not defined in real numbers}$$

But this one is $-\infty$ $$\lim_{x \to 2} \frac{-1}{(x-2)^2} = -\infty$$

Both denominator = 0 but different result.. I don't understand the difference..

The limits from the left and right differ (one is $\infty$, the other is $-\infty$). — David Mitra, Dec 23 '13 at 10:09
This is odd. Neither limit is defined in the real numbers. The first is not defined in the extended real numbers ($\Bbb R$ together with $\infty$ and $-\infty$), the second is defined in the extended reals. — David Mitra, Dec 23 '13 at 10:13

score 4 · Accepted Answer · answered Dec 23 '13 at 10:16

4

You may be misunderstanding limit.

The followings are correct.

$$\lim_{x\to 2-}\frac{x-3}{(x-2)(x+2)}=\infty$$ $$\lim_{x\to 2+}\frac{x-3}{(x-2)(x+2)}=-\infty$$ $$\lim_{x\to 2-} \frac{-1}{(x-2)^2}=-\infty$$ $$\lim_{x\to 2+}\frac{-1}{(x-2)^2}=-\infty$$

answered Dec 23 '13 at 10:16

mathlove

139,939

Well that makes much more sense. I did check the right and left limit. So if they differ then it is not defined I guess? – Chris Dec 23 '13 at 10:18
@Chris Yes, exactly. For the limit to be defined, the left-handed limit must be equal to the right-handed limit. – Ahaan S. Rungta Dec 23 '13 at 10:20
Ok. That explained it clearly enough then.. Finally this makes sense. Thanks a lot – Chris Dec 23 '13 at 10:22

score 4 · Answer 2 · answered Dec 23 '13 at 10:16

4

Case 1: $$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)}$$ $$= \frac{2+h-3}{(2+h-2)(2+h+2)}$$ $$= \frac{h-1}{(h)(4+h)} $$ $$= -\infty$$ Case 2: $$\lim_{x \to 2} \frac{x-3}{(x-2)(x+2)}$$ $$= \frac{2-h-3}{(2-h-2)(2-h+2)}$$ $$= \frac{-h-1}{(-h)(4-h)} $$ $$= +\infty$$

Hence undefined.

Where as for the other example both the limits are same and are equal to $-\infty$.Hence defined in that case.

answered Dec 23 '13 at 10:16

lsp

4,745

I inly admire your attempt for this limit. +1 – Mikasa Dec 23 '13 at 11:31

score 2 · Answer 3 · answered Dec 23 '13 at 10:20

2

Let's do what @David pointed.

When $x\to 2^+$ then $$\frac{1}{x-2}\to +\infty$$ and while $x\to 2^-$ then $$\frac{1}{x-2}\to -\infty$$ If you are not satisfied by these, take a very small value for $x$ from both sides of $2$. For $x\in 2^+$ set $x=2+ 10^{-100}$, so $$\frac{1}{x-2}=10^{100}$$ and if we set $x=2-10^{-100}$ then we have $$\frac{1}{x-2}=-10^{100}$$

answered Dec 23 '13 at 10:20

Mikasa

67,374

@amWhy: Good time my dear friend. I hope you had a nice slumber. :) – Mikasa Dec 23 '13 at 12:30
@amWhy: Great! I am well. I am writing some notes for undergraduate students. Maybe it becomes a book for them. :) – Mikasa Dec 23 '13 at 12:34
@SamiBenRomdhane: But it should be translated but the formulas. :D – Mikasa Dec 23 '13 at 17:22

score 1 · Answer 4 · answered Dec 23 '13 at 10:14

1

DavidMitra is exactly right. To see this, try graphing the two functions.

First one:

Second one:

answered Dec 23 '13 at 10:14

Ahaan S. Rungta

7,706

From the graph it is obvious that the second one goes to $- \infty$ but how do you find that out by the numbers? – Chris Dec 23 '13 at 10:17
First, note that this is equivalent to finding the limit $$ \lim_{x \to 0} \left( - \dfrac {1}{x^2} \right) $$ and then note that this is $0$. – Ahaan S. Rungta Dec 23 '13 at 10:21

Limit when denominator = 0

4 Answers4