Values of d where limit exists. lim x->d

  • #1
15
1

Homework Statement


For which value of d does the following limit exist?

lim x->d ln [ (x2-13x+30) / (x-d) ]

Homework Equations


None

The Attempt at a Solution


I understand how to find limits when the limit goes to a real number, and has a variable in the function to solve for, but not when the limit goes to a variable.

I thought perhaps attempting to multiply the whole thing by (x+d)/(x+d)

For simplicity I figure we can ignore that natural log for now.

(x-10)(x-3)(x+d)/(x-d)(x+d)

= (dx2 -13dx +30d +x3 -12x2 +30x) / (x2 + d2)

Which when simplified down gave me [ (-12x+47) / d ]

How do I go from here to solve for d though?

I tried subbing x as d, like I would solving for a limit normally, however this is on a program that provides instant feedback and ln(35) is incorrect.

Rather than the answer, as this is an assignment problem I'd much prefer direction or instruction on how to do this problem in general.
 

Answers and Replies

  • #2
13,477
10,530
Before bothering the limit, I would find out, which constellations of ##(x,d)## are possible under the logarithm, i.e. lead to a positive quotient.
Then the question is: Are different limits from the left and the right allowed or do the have to be the same? What if one exists and the other one doesn't?
 
  • #3
15
1
Before bothering the limit, I would find out, which constellations of ##(x,d)## are possible under the logarithm, i.e. lead to a positive quotient.
Then the question is: Are different limits from the left and the right allowed or do the have to be the same? What if one exists and the other one doesn't?
Okay, so a logarithm function is only defined for real, positive vales of x. In addition, while one sided limits do exist, the log function is a continuous function so if the LH limit exist, the RH limit exists, and in the case, the limits must be equal.
 
  • #4
13,477
10,530
Okay, so a logarithm function is only defined for real, positive vales of x. In addition, while one sided limits do exist, the log function is a continuous function so if the LH limit exist, the RH limit exists, and in the case, the limits must be equal.
Yes, but the function in the logarithm isn't continuous. What if you can approach from one side, while an approach from the other isn't defined? Anyway, since it is the logarithm of a number that goes to infinity, do you expect it to be finite somewhere?
 
  • #5
15
1
Yes, but the function in the logarithm isn't continuous. What if you can approach from one side, while an approach from the other isn't defined? Anyway, since it is the logarithm of a number that goes to infinity, do you expect it to be finite somewhere?
Okay, I can't answer that with my current knowledge. Before I go on, would you happen to be aware of some reading that would explain this situation to me?
 
  • #6
13,477
10,530
What do you mean? At ##x=d \pm \text{ something small }## we have a stable number in the nominator (##d^2-13d+30##) and a decreasing number in the denominator. Thus the quotient gets bigger and bigger, and the logarithm doesn't change that situation.
 
  • #7
15
1
What do you mean? At ##x=d \pm \text{ something small }## we have a stable number in the nominator (##d^2-13d+30##) and a decreasing number in the denominator. Thus the quotient gets bigger and bigger, and the logarithm doesn't change that situation.
Okay, so d must be limited to a small number because we don't want it to go to infinity (otherwise we wouldn't be able to find the value for d). Also, d must be positive because a negative log isn't defined and we don't have a value for d in the numerator to make the overall quotient positive?

Also, I meant reading like something to explain behaviour of logarithmic functions, or limits of logarithmic functions.
 
  • #8
15
1
OOOOOKKKKK.

So, ln[(x-10)(x-3)/(x-d)] has to be defined for the limit to exist.

We want to chose a value of d such that the denominator cancels part of the numerator.

If d=-3, then we have lim x-> -3 ln(x-10)
= ln(-13), which is undefined.

So try d=10
lim x->10 ln(x-3)
=ln(7)
which is defined.

So it's about analysing the problem to see what values are possible to remove the denominator and then which of those values lead to a defined answer.
 
  • Like
Likes fresh_42
  • #9
13,477
10,530
Okay, so d must be limited to a small number because we don't want it to go to infinity (otherwise we wouldn't be able to find the value for d).
That's a bit confusing. ##d## can be arbitrary, small or big. ##x-d## in the denominator, however, comes closer to zero as ##x## approaches ##d##. So ##\frac{1}{x-d}## gets to infinity, negative or positive depends on the side. It's the same as ##z \mapsto \frac{1}{z}## behaves at zero. The multiplication by the nominator by a finite number around ##d^2-13d+30## cannot compensate for the infinity we already have.
Also, d must be positive because a negative log isn't defined and we don't have a value for d in the numerator to make the overall quotient positive?
The entire quotient ##\frac{x^2-13x+30}{x-d}## must be positive. ##d## alone isn't important.
Also, I meant reading like something to explain behaviour of logarithmic functions, or limits of logarithmic functions.
You could read the Wikipedia page on logarithms https://en.wikipedia.org/wiki/Logarithm or play a little bit with Wolfram http://www.wolframalpha.com/input/?i=f(x)=lim(x->-2)log((x^2-13x+30)/(x+2))

But all this is rather heuristic. I guess you're supposed to put it in a formal proof.

Edit: You're right, I've forgotten the two possible cancellations.
 
  • Like
Likes ManicPIxie

Related Threads on Values of d where limit exists. lim x->d

Replies
1
Views
4K
Replies
9
Views
253
Replies
14
Views
3K
Replies
6
Views
2K
Replies
4
Views
8K
Replies
5
Views
2K
Replies
11
Views
2K
Replies
2
Views
10K
  • Last Post
Replies
6
Views
3K
Top