Having trouble evaluating these limits

[Nitrate]

Homework Statement

The limit as x approaches 2^+ of (x-4)/(x-2)

The limit as x approaches 2^- of (x-4)/(x-2)

Homework Equations

The Attempt at a Solution

Figured out that the domain is x ≠ 2, therefore 2 is a vertical asymptote
not sure where to input values/what to do really..

 

[murmillo]

drawing a picture will help, i think.

 

[Nitrate]

drawing a picture will help, i think.

I tried
lim
x→2⁺ (0⁺-4)/(0⁺-2)
but i keep getting + infinity

and I'm not even sure if that's how you approach this question.

Can anyone help?

 

[Mark44]

drawing a picture will help, i think.

I tried
lim
x→2⁺ (0⁺-4)/(0⁺-2)
but i keep getting + infinity

and I'm not even sure if that's how you approach this question.

Did you follow murmillo's suggestion?

As x approaches 2 from the right, what does the graph of y = (x - 4)/(x - 2) do?
As x approaches 2 from the left, what does the graph of y = (x - 4)/(x - 2) do?

 

[Nitrate]

Did you follow murmillo's suggestion?

As x approaches 2 from the right, what does the graph of y = (x - 4)/(x - 2) do?
As x approaches 2 from the left, what does the graph of y = (x - 4)/(x - 2) do?

I did not because I want to follow the method that my instructor used
i did however draw the asymptote

 

[SammyS]

I did not because I want to follow the method that my instructor used
i did however draw the asymptote

And what method is that?

 

[Nitrate]

And what method is that?

That's what I'm trying to piece together.
there was something about 1. (+)/(0^-) = - infinity
2. (-)/(0^-1) = + infinity
3. (-2)/(0^+)= - infinity
4. (1)/(0^+) = infinity

it has worked so far for my previous examples such as:
lim
x→1⁺ (3/x-1)

Domain: x can not equal 1
V.A: x = 1

lim (3)/(0⁺-1) = + infinity

but it is not working for the example that I'm on and i don't know why

 

[SammyS]

Homework Statement

The limit as x approaches 2^+ of (x-4)/(x-2)

The limit as x approaches 2^- of (x-4)/(x-2)

Homework Equations

The Attempt at a Solution

Figured out that the domain is x ≠ 2, therefore 2 is a vertical asymptote
not sure where to input values/what to do really..

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

As x approaches 2 from the right (from the positive x direction) x-2 → 0⁺, that is to say x-2 becomes very small, but is positive.

Also, as x approaches 2 from the right, x-4 is close to positive 2.

So this gives you (+2)/(0⁺) → +∞​

The limit as x → 2^- can be handled similarly.

 

[Nitrate]

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

As x approaches 2 from the right (from the positive x direction) x-2 → 0⁺, that is to say x-2 becomes very small, but is positive.

Also, as x approaches 2 from the right, x-4 is close to positive 2.

So this gives you (+2)/(0⁺) → +∞​

The limit as x → 2^- can be handled similarly.

however, for the first limit, the answer in the back of the book comes up as -∞
and when i type the limit as x approaches 2^+ of (x-4)/(x-2) into wolfram alpha it also comes up as - infinity
http://www.wolframalpha.com/input/?i=the+limit+as+x+approaches+2^++of+(x-4)/(x-2)

 

[Nitrate]

can anyone else help?

 

[SammyS]

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

As x approaches 2 from the right (from the positive x direction) x-2 → 0⁺, that is to say x-2 becomes very small, but is positive.

Also, as x approaches 2 from the right, x-4 is close to positive 2.

So this gives you (+2)/(0⁺) → +∞​

The limit as x → 2^- can be handled similarly.

DUH !

I made a mistake !

... , as x approaches 2 from the right, x-4 is close to -2 (negative 2).

Therefore, this gives you (-2)/(0⁺) → -∞

 

[Nitrate]

DUH !

I made a mistake !

... , as x approaches 2 from the right, x-4 is close to -2 (negative 2).

Therefore, this gives you (-2)/(0⁺) → -∞

sorry, but I'm having trouble understanding why x-4 is close to -2

 

[Mark44]

sorry, but I'm having trouble understanding why x-4 is close to -2

The whole statement is, when x is close to 2, x - 4 is close to -2.

 

[Mark44]

I did not because I want to follow the method that my instructor used
i did however draw the asymptote

You should still sketch a graph of the function. The intent wasn't meant to be a proof, but to enable you to have some insight as to what the limit should be.

 

[Nitrate]

finally got the concept!
thanks everyone :)