Is the following question legitimate/sensible?

[Agent Smith]

Find \(\displaystyle \displaystyle \lim_{x \to a} | \log x - a |\)

If it is how would we solve it?

It popped up when I was graphing \(\displaystyle |\log x - x| < 0.8\) as part of my attempt to finding a number which is its own logarithm (in this case base 10, but some might prefer the more "natural" Euler number, e)

 

[Dr.Peterson]

Find \(\displaystyle \displaystyle \lim_{x \to a} | \log x - a |\)

If it is how would we solve it?

It popped up when I was graphing \(\displaystyle |\log x - x| < 0.8\) as part of my attempt to finding a number which is its own logarithm (in this case base 10, but some might prefer the more "natural" Euler number, e)

It's a legitimate question, but not interesting. Since the log is continuous, the limit is just |log(a) - a|, whatever that might be. If a = 10, for example, it's just |1 - 10| = 9.

 

[mario99]

Find \(\displaystyle \displaystyle \lim_{x \to a} | \log x - a |\)

Do you mean [imath]\log(x - a)[/imath]?

 

[Agent Smith]

Do you mean [imath]\log(x - a)[/imath]?

No, non cogito no, I mean |log x - a|. Is it an awkward expression? There's nothing new about it looks like because there's an answer. I forgot the specifics, nevertheless it ends something like \(\displaystyle e = x^{1/x}\), for base e.

 

[Agent Smith]

It's a legitimate question, but not interesting. Since the log is continuous, the limit is just |log(a) - a|, whatever that might be. If a = 10, for example, it's just |1 - 10| = 9.

Sad, how do we make it interesting? I was tinkering around with \(\displaystyle 2^x = x\). No solution (per graph, though there's a "closest approach" at x = 1 (?)).

Doesn't, shouldn't \(\displaystyle x = a \pm h\) be the standard approach to compute the limit?

 

[mario99]

No, non cogito no, I mean |log x - a|. Is it an awkward expression? There's nothing new about it looks like because there's an answer. I forgot the specifics, nevertheless it ends something like \(\displaystyle e = x^{1/x}\), for base e.

It is not an awkward expression, but it leads to different calculations. It is better to write it as [imath]\displaystyle \log(x) - a[/imath].

Or

[imath]\displaystyle -\left(a - \log x\right)[/imath]

Doing this will make it clear what is the parameter of the logarithmic function.

 

[Agent Smith]

It is not an awkward expression, but it leads to different calculations. It is better to write it as [imath]\displaystyle \log(x) - a[/imath].

Or

[imath]\displaystyle -\left(a - \log x\right)[/imath]

Doing this will make it clear what is the parameter of the logarithmic function.

Ok. Thanks

 

[mario99]

Ok. Thanks

You are welcome.

But I am sad because the limit was not [imath]\displaystyle \lim_{x\rightarrow a}|\ln(x - a)|[/imath].

That would be an interesting problem!

 

[Agent Smith]

Doesn't, shouldn't x=a±h be the standard approach to compute the limit?

@mario99

 

[mario99]

@mario99

No. The standard approach to compute the limit is to plug in the value, say [imath]a[/imath], directly. If you get a weird result, you try to investigate by various methods. One of them is to check what happens at [imath]a \pm h[/imath].

You are so lucky Agent Smith as your limit is very innocent which means no investigation is needed.

[imath]\displaystyle \lim_{x\rightarrow a} \big|\log(x) - a\big| = \big|\log(a) - a\big| \ \ \ [/imath] which is a perfect result.

 

[Agent Smith]

No. The standard approach to compute the limit is to plug in the value, say [imath]a[/imath], directly. If you get a weird result, you try to investigate by various methods. One of them is to check what happens at [imath]a \pm h[/imath].

You are so lucky Agent Smith as your limit is very innocent which means no investigation is needed.

[imath]\displaystyle \lim_{x\rightarrow a} \big|\log(x) - a\big| = \big|\log(a) - a\big| \ \ \ [/imath] which is a perfect result.

Can I construct a table for the function around a?

 

[mario99]

Can I construct a table for the function around a?

Why do you want to do that?

 

[Agent Smith]

Why do you want to do that?

It's taught as a modus of finding the limit, oui?

 

[mario99]

Can I construct a table for the function around a?

The answer to your question is yes you can.

It's taught as a modus of finding the limit, oui?

But logarithmic functions are not very nice with negative numbers. And because of that you don't want to construct a table around [imath]a[/imath].

A table around [imath]a[/imath] needs complex analysis which is beyond the scope of this thread.

 

[Agent Smith]

The answer to your question is yes you can.


But logarithmic functions are not very nice with negative numbers. And because of that you don't want to construct a table around [imath]a[/imath].

A table around [imath]a[/imath] needs complex analysis which is beyond the scope of this thread.

but a doesn't have to be negative, oui?

 

[mario99]

but a doesn't have to be negative, oui?

In this case, we can choose [imath]a = 1[/imath]. And we know that [imath]\ln 1 = 0[/imath]. Our goal is constract a table around [imath]a = 1[/imath] to show that [imath]\ln 1 = 0[/imath].

This means we need to find:
[imath]\ln 0.9[/imath]
[imath]\ln 0.99[/imath]
[imath]\ln 0.999[/imath]
[imath]\ln 0.9999[/imath]
And
[imath]\ln 1.1[/imath]
[imath]\ln 1.01[/imath]
[imath]\ln 1.001[/imath]
[imath]\ln 1.0001[/imath]

Try it. What do you get?

If you want to be fancy, and not to use the function [imath]\ln x[/imath] to get the values, you can use Taylor series centered at [imath]x = 1[/imath].

[imath]\displaystyle \ln x = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}(x - 1)^n[/imath]

For example,

[imath]\displaystyle \ln 0.9999 = \sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n}(0.9999 - 1)^n[/imath]

If you think that infinite series is beyond your knowledge, you can use linear approximation.

[imath]\displaystyle \ln x \approx \ln x_0 + \frac{1}{x_0}\Delta x[/imath]

For example,

[imath]\displaystyle \ln 0.9999 \approx \ln 1 + \frac{1}{1}(0.9999 - 1)[/imath]