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Delta-epsilon limit of a rational function with a jump discontinuity

  • Thread starter NickBE
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  • #1
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Hello, I am trying to prove the following...

lim (x+3) [tex]\left|x+5\right|[/tex][tex]/x+5[/tex]
x[tex]\rightarrow-5[/tex]

from the left, L=+2
from the right, L=-2

I used delta-epsilon on the right hand limit and got [tex]\delta[/tex] = [tex]\epsilon[/tex]

However, I'm not sure how to proceed when I get to this step while trying to prove the left hand limit:

[tex]\left|x+1\right|[/tex]< [tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex]<[tex]\delta[/tex]
 

Answers and Replies

  • #2
Dick
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If you are approaching from the left then |x+5|/(x+5)=(-1). So your expression f(x) becomes (-1)*(x+3). The quantity you want to prove less than epsilon is |L-f(x)|. I make that to be |2-(-1)(x+3)|=|x+5|.
 
  • #3
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I think I see where you're going, but I want to make sure I have the right idea with the delta-epsilon definition.

lim f(x)=L
x->a

[tex]\left|f(x)-L\right|[/tex] <[tex]\epsilon[/tex] if 0<[tex]\left|x-a\right|[/tex]< [tex]\delta[/tex]

Is this the same definition that is used to find the one sided limits? If so, how is it manipulated to arrive at the conclusion you provided?

What I was trying to do as I attempted to prove the limit from the right:

[tex]\left|f(x)+2\right|[/tex] < [tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex] < [tex]\delta[/tex]

Which gave me:

[tex]\left|x+5\right|[/tex] <[tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex] < [tex]\delta[/tex]

and I called it a day on that one. But when I got to the second one, I tried using the same definition to try to prove the limit. I have a feeling that in the same format, it will give me an incorrect answer...where I am lacking understanding?
 

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