Proving a Limit: Epsilon & Delta Solution

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[SOLVED] Proving a limit

\lim_{x \rightarrow 5} \frac{x^3-6x^2+15x-47}{\{x^2-8x+14}=-3}

So I know how to prove a limit using epsilon and delta. The only thing is my book doesn't show how to do it with a fraction. So i get stuck at a certain point.

I get up to \frac{x-5\times (x^2+2x+1)}{\{x^2-8x+14}\leq\epsilon}

But then what do i do with the numerator {(x^2+2x+1)}

and the denominator {(x^2-8x+14)}

Also I' am not sure if i restrict x to \frac{9}{\{2}\leq x \leq\frac{11}{\{2}

Basically I need someone to point me in the right direction about what to do with the numerator and denominator. Thanks for any help given.

P.S : Ignore the symbol in front of the x^2 and the 2 in the denominator( I didnt know how to get rid of it). I also didnt put the absolute value where they go because i didn't know how but hopefully you guys understand. If not then thanks anyway
 
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If I posted something wrong or if it's not clear just tell me and i will try and change it so it makes sense.
 
Been a while since I've done epsilon delta proofs... but I'll try to help. you want |f(x) - L|<epsilon...

what expression do you get for |f(x) - L| ?
 
thanks for the reply. Well i did the f(x)-L. What i got was the fraction i showed above. This is after i combined them and factored out the x-5 to get the (x- a) is less than or equal to delta. Basically the thing below. What I don't know is what to do after this.

\frac{x-5\times (x^2+2x+1)}{\{x^2-8x+14}\leq\epsilon}
 
Oops... never mind... let me have a look. sorry.
 
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Yes, looks like you're almost there... I think the trick is to find the maximum value of |\frac{x^2+2x+1}{x^2-8x+14}|... or if we can say it is bounded by some maximum value then you can finish...

Not sure how exactly...
 
The trick is to pick a temporary delta neighborhood (remember your temporary delta) and bound your function. Bound the numerator and denominator separately, remembering the archemidian principle for the denominator. Youll come up delta being less than epsilon times a coefficient as usual. But you have to cover that the temporary delta inequality from the start is always secured regardless of epsilon. So your choice of delta winds up being the minimum of a number and a function of epsilon (hope this makes some sense)
 
I think if restrict the x to 9/2 -11/2 interval because i think i need it to be below the graph and above the x- axis. So Then if i choose 9/2 as the value and what i get is 30.25 over 1.75. So what i think the answer is...

\delta= min(\frac{1}2},{.057377\epsilon)

But I'm not sure if it's correct and can't quite prove it
 
turdferguson said:
The trick is to pick a temporary delta neighborhood (remember your temporary delta) and bound your function. Bound the numerator and denominator separately, remembering the archemidian principle for the denominator. Youll come up delta being less than epsilon times a coefficient as usual. But you have to cover that the temporary delta inequality from the start is always secured regardless of epsilon. So your choice of delta winds up being the minimum of a number and a function of epsilon (hope this makes some sense)

Sorry I never learned this. All i know is to restrict x to an interval and then choose (depending on the problem) the higher number and then plug it into x to get the epsilon. My Teacher doesn't explain things to clearly and it's not in my textbook so I'm at a loss...


Edit: I think it's the same thing as I did up top but once again I'm not really sure. So any help would be appreciated
 
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  • #10
Hmmm... ok we can restrict x as much as we want...

so suppose we restrict |x-5|<1/2

ie 4.5<x<5.5...

from your equation:

|x-5|\le\frac{\epsilon}{|\frac{x^2+2x+1}{x^2-8x+14}|}

so we need the maximum value of |\frac{x^2+2x+1}{x^2-8x+14}|... from
4.5<x<5.5, we can choose the maximum of the numerator magnitude wise divided the minimum of the denominator magnitude wise... and we'll know that the maximum of |\frac{x^2+2x+1}{x^2-8x+14}| is less than that...
 
  • #11
learningphysics said:
Hmmm... ok we can restrict x as much as we want...

so suppose we restrict |x-5|<1/2

ie 4.5<x<5.5...

from your equation:

|x-5|\le\frac{\epsilon}{|\frac{x^2+2x+1}{x^2-8x+14}|}

so we need the maximum value of |\frac{x^2+2x+1}{x^2-8x+14}|... from
4.5<x<5.5, we can choose the maximum of the numerator magnitude wise divided the minimum of the denominator magnitude wise... and we'll know that the maximum of |\frac{x^2+2x+1}{x^2-8x+14}| is less than that...

This is what i did but i chose the minimum, meaning 4.5 because i thought that we need lower interval. So what i did then was substitute the 4.5 in the numerator and denominator. Then i got the answer was around .057377 times epsilon


So if i understand you correctly i choose max for numerator and min for the denominator which makes more sense. But then how do i prove it?
 
  • #12
Alright when i solve the equation |x-5|\le\frac{\epsilon}{|\frac{x^2+2x+1}{x^2-8x+14}|}

The way you said i get |x-5|\le\\epsilon\times.04142}}

which i believe is right. Now i just don't understand how to prove it. Suppose i substitute the epsilon times .04142 as delta. How would i go about to show the original equation.

P.S. : I hope you understand what I mean at least sort of. Basically how would i prove this is in fact the answer
 
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  • #13
So for maximum of the numerator I get |(5.5)^2 + 2(5.5) + 1| = 42.25

For the minimum of the denominator I get |(4.5)^2 - 8(4.5) + 14| = 1.75

I get delta as 0.0414epsilon...

I think any interval that doesn't include the zeros of x^2-8x+14 would be fine...

Anyway, to prove that the limit is -3, you have to work the other way right? Given an epsilon > 0. show that for delta = min{1/2, 0.0414epsilon} where |x-5|<delta, |f(x) + 3|<epsilon.

So start with |f(x) + 3|... and show that for |x-5|<delta for the given delta... we get |f(x)+3|<epsilon

so you'll have two cases... 1/2 <=0.0414epsilon, so delta =1/2 in this case... and in the second case 1/2>0.414epsilon and delta =0.414epsilon.
 
  • #14
Actually, I think both cases can be handled together...
 
  • #15
learningphysics said:
So for maximum of the numerator I get |(5.5)^2 + 2(5.5) + 1| = 42.25

For the minimum of the denominator I get |(4.5)^2 - 8(4.5) + 14| = 1.75

I get delta as 0.0414epsilon...

I think any interval that doesn't include the zeros of x^2-8x+14 would be fine...

Anyway, to prove that the limit is -3, you have to work the other way right? Given an epsilon > 0. show that for delta = min{1/2, 0.0414epsilon} where |x-5|<delta, |f(x) + 3|<epsilon.

So start with |f(x) + 3|... and show that for |x-5|<delta for the given delta... we get |f(x)+3|<epsilon

so you'll have two cases... 1/2 <=0.0414epsilon, so delta =1/2 in this case... and in the second case 1/2>0.414epsilon and delta =0.414epsilon.

Alright the part i still don't understand is when i plug in the delta 1/2 into |x-5|<delta how do we get it to the |f(x)+3|<epsilon
 
  • #16
SO when i plug in 1/2 as delta. |x-5|< 1/2 how do i show that it's the same as |f(x)+3|<epsilon
 
  • #17
we're given an epsilon... we choose delta = min{1/2, 0.0414epsilon} now we want to show that:

|\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3| is less than epsilon when |x-5|<delta, for our delta.

this equals

|\frac{(x-5)(x^2+2x+1)}{x^2-8x+14}|

we know that for |x-5|<delta... where delta = min{1/2, 0.0414epsilon}, |x-5|<1/2 (since delta is always less than or equal to 1/2) AND |x-5|<0.0414epsilon for the same reason.

we also know that the maximum value of |\frac{x^2+2x+1}{x^2-8x+14}| is less than (42.25/1.75) as we showed before (we showed it true for |x-5|<1/2... now we either have the same interval or a smaller one so the maximum value is either the same or smaller)...

so |\frac{(x-5)(x^2+2x+1)}{x^2-8x+14}| < |x-5|*(max vale of |\frac{x^2+2x+1}{x^2-8x+14}|) < |x-5|(42.25/1.75)

we know that |x-5|<0.0414epsilon, so |x-5|(42.25/1.75) < 0.0414epsilon(42.25/1.75)<epsilon
 
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  • #18
might be better to use (1.75/42.25 )epsilon instead of 0.0414epsilon... just cleaner to use exact numbers...
 
  • #19
i still don't get what i have to do for the proof. I know we have to work backwards but when we substitute the x for the max and min we get rid of the equation so i don't see how we get it if we go backwards.

From |x-5|<0.0414epsilon i go to |x-5|(42.25/1.75)< epsilon

but then how do i get it to look like the original equation |\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3| is less than epsilon
 
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  • #20
ace123 said:
i still don't get what i have to do for the proof

I proved it in post #17... can you describe which part(s) are confusing... how do you normally do epsilon delta proofs?
 
  • #21
what i mean is for example
lim(4x-3)=5
x->2

where s is going to equal e/4

then to prove it i normally plug in e/4 as s...

x-2<e/4
4(x-2)<e
4x-8<e which is our original f(x)-L

This is how I did it for other limits
 
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  • #22
ace123 said:
i still don't get what i have to do for the proof. I know we have to work backwards but when we substitute the x for the max and min we get rid of the equation so i don't see how we get it if we go backwards.

From |x-5|<0.0414epsilon i go to |x-5|(42.25/1.75)< epsilon

but then how do i get it to look like the original equation |\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3| is less than epsilon

before that part, I wrote:

"so |\frac{(x-5)(x^2+2x+1)}{x^2-8x+14}| < |x-5|*(max value of |\frac{x^2+2x+1}{x^2-8x+14}|) < |x-5|(42.25/1.75)"
 
  • #23
learningphysics said:
before that part, I wrote:

"so |\frac{(x-5)(x^2+2x+1)}{x^2-8x+14}| < |x-5|*(max value of |\frac{x^2+2x+1}{x^2-8x+14}|) < |x-5|(42.25/1.75)"

yea so i don't understand how that works. What do you mean for max value. The 5.5? and what do i do plug it in?
 
  • #24
Oh.. I miss understood your max value. So i understand how to prove the 42.25/1.75 but still unsure about the 1/2. Since their is no epsilon what do i do after

|x-5|< 1/2 to get it to once again the original equation |\frac{x^3-6x^2+15x-47}{x^2-8x+14}+3| is less than epsilon
 
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  • #25
ace123 said:
yea so i don't understand how that works. What do you mean for max value. The 5.5? and what do i do plug it in?

Yeah, remember when we plugged in 5.5 for the numerator, and 4.5 for the denominator for that equation... we were finding a bound for that function... ie: we were showing that that function will take a value less than 42.5/1.75... We took the maximum value for the numerator... minimum for the denominator... over |x-5|<1/2.

we're just using that same bound again...
 
  • #26
ace123 said:
what i mean is for example
lim(4x-3)=5
x->2

where s is going to equal e/4

then to prove it i normally plug in e/4 as s...

x-2<e/4
4(x-2)<e
4x-8<e which is our original f(x)-L

This is how I did it for other limits

Oh i see... you can do it this way then

|x-5|<delta

Then multiply the left side by |(x^2+2x+1)/(x^2-8x+14)|, and multiply the right side by 42.5/1.75

That gives us:

|(x-5)(x^2+2x+1)/(x^2-8x+14)|<delta*(42.5/1.75) (the inequality holds because 42.5/1.75 is greater than the max value)

we know that delta<=(1.75/42.5)epsilon

so delta*(42.5/1.75)<=epsilon

so we get:

|(x-5)(x^2+2x+1)/(x^2-8x+14)|<epsilon

:redface: yeah, that's a much better way to do it.
 
  • #27
Okay great i understand the epsilon proof much better. But how would i go about proving 1/2? Since their is no epsilon there
 
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  • #28
ace123 said:
Okay great i understand the epsilon proof much better. But how would i go about proving 1/2? Since their is no epsilon there

we did it together, at the same time...

since |x-5|<delta, we know that |x-5|<1/2 and |x-5|<(1.75/42.25)epsilon (no matter what the actual delta value is, both these conditions are true...)

when we multiplied by the max value 42.25/1.75, we were using the fact that |x-5|<1/2.

so we proved it for both cases.
 
  • #29
I'm not going to lie to you. I don't see how when we multiply by 42.25/1.75, we were using the fact that |x-5|<1/2.

But i guess it doesn't really matter. If you say it proves it for both then it must be so.

Anyway thanks for all the help you gave me with this problem. And more importantly I'll be thinking of you and your help as I can now ace my calculus exam. Your my hero!
 
  • #30
ace123 said:
I'm not going to lie to you. I don't see how when we multiply by 42.25/1.75, we were using the fact that |x-5|<1/2.

But i guess it doesn't really matter. If you say it proves it for both then it must be so.

Anyway thanks for all the help you gave me with this problem. And more importantly I'll be thinking of you and your help as I can now ace my calculus exam. Your my hero!

|(x^2+2x+1)/(x^2-8x+14)| is bounded above by 42.25/1.75 over the range of |x-5|<1/2. If we didn't know that |x-5|<1/2, we wouldn't be able to use this bound.

Thanks. you're welcome. I really appreciate the compliment. :smile: But don't take my word for anything, or anyone else... convince yourself... it's possible I made a mistake. anyway, good luck on your exam!
 
  • #31
learningphysics said:
|(x^2+2x+1)/(x^2-8x+14)| is bounded above by 42.25/1.75 over the range of |x-5|<1/2. If we didn't know that |x-5|<1/2, we wouldn't be able to use this bound.

Thanks. you're welcome. I really appreciate the compliment. :smile: But don't take my word for anything, or anyone else... convince yourself... it's possible I made a mistake. anyway, good luck on your exam!

Oh I understand now. I guess i completely forgot that we had to restrict x to the 1/2 interval. It makes complete sense now. I really appreciate you help and thanks :smile:

P.S. Wish you were my Professor, then life would be so much simplier
 

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