# Is the following question legitimate/sensible?

#### Agent Smith

##### Full Member
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)

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.

Find $$\displaystyle \displaystyle \lim_{x \to a} | \log x - a |$$
Do you mean [imath]\log(x - a)[/imath]?

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.

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?

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.

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

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!

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.

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?

Can I construct a table for the function around a?
Why do you want to do that?

Why do you want to do that?

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

Can I construct a table for the function around a?

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.

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?

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]