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

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In summary: The (1-ln(x)/x) part is what goes to 1 as x goes to infinity.In summary, the conversation discusses different approaches to finding the value of a limit as x approaches infinity. One person suggests using L'Hopital's rule, while another suggests looking at the derivative of the function. They also discuss the role of ln(x) in the expression and the value of ln(e). Ultimately, it is determined that the limit is equal to infinity.
  • #1
r_swayze
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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 -∞
 
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  • #2
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.
 
  • #3
Why don't you apply l'Hopital to the ln(x)/x part of your expression and use the result to simplify it?
 
  • #4
(At the risk of confusing things) or take the first x into the ln(x).
 
  • #5
Remember that [tex] ln(x) = \int_1^x \frac{1}{u} du [/tex] and use the MVT
 
  • #6
Dick said:
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?
 
  • #7
This is what I would do:

[tex]\stackrel{lim}{x\rightarrow\infty}(x-ln(x))\times\frac{(x+ln(x))}{(x+ln(x))}[/tex]

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

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

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.
 
  • #8
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.
 
  • #9
zcd said:
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?
 
  • #10
Yes, ln e= 1.
 
  • #11
r_swayze said:
where does the ln(x)/x come from?

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

1. What is the purpose of solving lim x→∞ (x - ln x)?

The purpose of solving this limit is to determine the behavior of the function as x approaches infinity. This can help in understanding the overall trend of the function and its growth rate.

2. What are the steps to solve lim x→∞ (x - ln x)?

The steps to solve this limit are as follows:
1. Rewrite the expression as a fraction with x in the denominator
2. Use L'Hôpital's rule to differentiate the numerator and denominator
3. Simplify the resulting expression
4. Take the limit as x approaches infinity
5. If necessary, use algebraic manipulation to simplify the expression further
6. The resulting value is the limit as x approaches infinity.

3. Can this limit be solved without using L'Hôpital's rule?

Yes, this limit can also be solved by using other techniques such as the squeeze theorem or using properties of limits. However, L'Hôpital's rule is often the simplest and most efficient method to use.

4. What is the significance of the natural logarithm in this limit?

The natural logarithm is used in this limit because it is the inverse function of the exponential function, which is commonly used to model growth and decay in natural phenomena. As x approaches infinity, the natural logarithm helps to determine the rate at which the function grows or decays.

5. How can solving this limit be applied in real-world scenarios?

Solving this limit can be applied in various real-world scenarios, such as in economics, physics, and biology. For example, in economics, this limit can be used to analyze the long-term growth of a company's profits. In physics, it can be used to study the motion of objects with changing velocity. In biology, it can be used to model the growth of populations or the decay of radioactive materials.

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