Solving Lim x→∞ (x - ln x): Step by Step Guide

[r_swayze]

Can somebody show a step by step process of how to get this answer?

I keep getting -∞ when it should be ∞

heres my work:

lim x→∞ (x - ln x) = lim x→∞ ((1 - 1/x ln x) / (1/x)

then apply l'hospitals rule

lim x→∞ (x^(-2) ln x - x^(-2)) / (-x^(-2))

then I cancel all the x^(-2) and I am left with -ln x which would equal to -∞

 

[Mark44]

You can apply L'Hopital's rule only when you have an indeterminate form of [0/0] or [+/-infinity/infinity].

In this expression -- ((1 - 1/x ln x) / (1/x) -- 1/x approaches zero, but ln x approaches infinity, which makes it another indeterminate form.

I don't think that there's anything you can do with L'Hopital's rule on this one. You might try an approach that looks at the derivative of f(x) = x - ln x. It's easy to show that the derivative approaches 1 as x gets large, which suggests that x - ln x gets larger and larger.

 

[Dick]

Why don't you apply l'Hopital to the ln(x)/x part of your expression and use the result to simplify it?

 

[George Jones]

(At the risk of confusing things) or take the first x into the ln(x).

 

[JG89]

Remember that $$ln(x) = \int_1^x \frac{1}{u} du$$ and use the MVT

 

[r_swayze]

Why don't you apply l'Hopital to the ln(x)/x part of your expression and use the result to simplify it?

where does the ln(x)/x come from?

 

[physicsnoob93]

This is what I would do:

$$\stackrel{lim}{x\rightarrow\infty}(x-ln(x))\times\frac{(x+ln(x))}{(x+ln(x))}$$

This would give:$$\stackrel{lim}{x\rightarrow\infty}\frac{(x^{2}-ln^{2}(x))}{(x+ln(x))}$$

You could work it out from there, split the limit up by $$x^{2}$$ and $$ln^{2}x$$

Hope this helped, and sorry about the poor formatting, this is probably the first time I am doing latex.

By the way, I'm sure the value is infinity.

 

[zcd]

You could do lim x→∞ (ln e^x - ln x) = lim x→∞ ln (e^x /x) and apply l'hopital's rule inside; the domain change of x to ln e^x is irrelevant because your limit exists within the domain of log.

 

[r_swayze]

You could do lim x→∞ (ln e^x - ln x) = lim x→∞ ln (e^x /x) and apply l'hopital's rule inside; the domain change of x to ln e^x is irrelevant because your limit exists within the domain of log.

I think I see it, but just to make sure, does ln e equal 1? so that can be plugged into any equation?

 

[HallsofIvy]

Yes, ln e= 1.

 

[Dick]

where does the ln(x)/x come from?

Write it (as you basically did) as x*(1-ln(x)/x).