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

• ManicPIxie
In summary, the homework equation for which limit exists is limited to a small number and must be positive.
ManicPIxie

## Homework Statement

For which value of d does the following limit exist?

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

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.

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?

fresh_42 said:
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.

ManicPIxie said:
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?

fresh_42 said:
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?

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.

fresh_42 said:
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.

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.

fresh_42
ManicPIxie said:
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.

ManicPIxie

## 1. What does it mean for a limit to exist at a certain value of d?

When a limit exists at a certain value of d, it means that as x approaches d from both sides, the function approaches a single, finite value. In other words, the function is continuous at that point and there are no gaps or jumps in its graph.

## 2. How can I determine if a limit exists at a given value of d?

To determine if a limit exists at a given value of d, you can use the following methods:

• Direct Substitution: If the function is defined at d, then the limit exists and is equal to the function value at d.
• Numerical Approximation: By plugging in values of x that are very close to d, you can get an idea of what the limit might be.
• Graphical Analysis: You can use a graphing calculator or software to graph the function and see if there are any gaps or holes at d.
• Algebraic Manipulation: Sometimes, you can manipulate the function algebraically to simplify it and determine the limit at d.

## 3. Can a limit exist at a value of d if the function is not defined at that point?

Yes, a limit can exist at a value of d even if the function is not defined at that point. This is because limits are concerned with the behavior of a function near a certain point, not necessarily at that point.

## 4. Is it possible for a limit to exist at a value of d, but for the function to not be continuous at that point?

Yes, it is possible for a limit to exist at a value of d, but for the function to not be continuous at that point. This can happen if the function has a removable discontinuity, such as a hole or jump at that point.

## 5. How can I use limits to find the value of d where a function is not continuous?

To find the value of d where a function is not continuous, you can use the following steps:

1. Check if the function is defined at d. If it is not, then the function is not continuous at d.
2. Check if the limit exists at d. If it does not, then the function is not continuous at d.
3. If the limit exists but is not equal to the function value at d, then the function has a removable discontinuity at d.
4. To find the value of d where the function is not continuous, set the limit equal to the function value and solve for d.

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