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Having trouble evaluating these limits

  • Thread starter Nitrate
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  • #1
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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..
 

Answers and Replies

  • #2
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drawing a picture will help, i think.
 
  • #3
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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?
 
Last edited:
  • #4
33,262
4,963
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?
 
  • #5
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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
 
  • #6
SammyS
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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?
 
  • #7
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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
 
  • #8
SammyS
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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 [itex]\displaystyle \lim_{x\to2^+} \frac{x-4}{x-2}[/itex]
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.
 
  • #9
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For [itex]\displaystyle \lim_{x\to2^+} \frac{x-4}{x-2}[/itex]
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)
 
  • #10
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can anyone else help?
 
  • #11
SammyS
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For [itex]\displaystyle \lim_{x\to2^+} \frac{x-4}{x-2}[/itex]
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+) → -∞
 
  • #12
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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
 
  • #13
33,262
4,963
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.
 
  • #14
33,262
4,963
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.
 
  • #15
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finally got the concept!
thanks everyone :)
 

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