Limit of x - lnx as x approaches infinity: Solving Indeterminate Difference"

[TsAmE]

Homework Statement

Find lim x -> ∞ (x - lnx)

Homework Equations

None.

The Attempt at a Solution

lim x -> ∞ (x - lnx)
= lim x -> ∞ ( 1 / 1/x ) - ( 1 / 1/lnx)
= lim x -> ∞ ( 1 / lnx ) - ( 1 / x) / (1 / xlnx)

My working out is just getting bigger and bigger and I don't know what to do :(. I do know that this is an indeterminate difference of the form ∞ - ∞

 

[Mentallic]

Have you learned L'Hospital's Rule?

If not you can just use logic to explain that lnx grows much more slowly than x.

$$lim(x-lnx)=lim(x)-lim(ln(x))=lim(x)-ln(lim(x))$$

Can you see why the limit is approaching infinite?

 

[TsAmE]

I tried using l'Hospital. I differentiated and simplified to get:

lim x -> ∞ ( -1 / x(lnx)^(2) + 1 / x^2 ) / ( -1 / (lnx)^2 + lnx ) and this ended up to be 0, but the correct answer is ∞?

 

[Mark44]

I tried using l'Hospital. I differentiated and simplified to get:

lim x -> ∞ ( -1 / x(lnx)^(2) + 1 / x^2 ) / ( -1 / (lnx)^2 + lnx ) and this ended up to be 0, but the correct answer is ∞?

Yes, but you used L'Hopital's Rule incorrectly. It can be used only on expressions that are quotients, which you don't have. If you work with your expression, you can turn it into a product, and then turn that into a quotient.

 

[TsAmE]

Yes, but you used L'Hopital's Rule incorrectly. It can be used only on expressions that are quotients, which you don't have. If you work with your expression, you can turn it into a product, and then turn that into a quotient.

I differentiated the numerator and denominator according to l'Hospital's rule so I didnt violate the rule I differentiated the numerator: ( 1 / lnx - 1 / x ) divided the derivative of denominator: ( 1 / xlnx ) to get to the answer in my previous post

 

[Mark44]

Is this what you're referring to?

lim x -> ∞ (x - lnx)
= lim x -> ∞ ( 1 / 1/x ) - ( 1 / 1/lnx)
= lim x -> ∞ ( 1 / lnx ) - ( 1 / x) / (1 / xlnx)

When I saw the above, I thought that you were using L'Hopital's Rule, but now I realize that you weren't. For the third line to be interpreted as a quotient it needs a pair of brackets or parentheses to clearly mark what is the numerator and what is the denominator, like so:

[( 1 / lnx ) - ( 1 / x)] / (1 / xlnx)

Now that I understand what you were trying to write, that is suitable for using L'Hopital's Rule, but with all those reciprocals, it looks like it would be very messy. Which might be why you ended up with a value of 0 instead of the right value, which is infinity.

It's much simpler to write x - lnx as x(1 - lnx/x), and note that 1 - lnx/x --> 1 as x --> infinity, so the whole product approaches infinity. I think that's a reasonable approach.

 

[TsAmE]

For lim x -> ∞ (x - lnx) using your form of x(1 - lnx / x) I got:

∞(1 - ∞ / ∞) which leads to an indeterminate type in the brackets.

 

[Mark44]

Right. So to evaluate lim [x(1 - lnx)/x] you can break it up, which is legitimate to do provided that the limits exist (in the broadest sense).

lim [x(1 - lnx)/x] = lim x * lim(1 - lnx/x) = lim x * [lim 1 - lim(lnx/x)], with all limits taken as x --> infinity. You can use L'Hopital's Rule on lim(lnx/x).

There might be a better way, but if there is, it hasn't occurred to me.

 

[Tedjn]

Here is another approach. It is easy to show that for x > e the function (1/e)x is greater than the function ln(x) by comparing derivatives. So, for x > e, x-(1/e)x < x-ln(x), which gives a lower bound for our function. This lower bound is unbounded, which is enough to show that our original function is unbounded.

 

[vela]

You could also use x=ln e^x and use the properties of the logarithm to combine the terms. The argument of the log will be an indeterminate form you can use the hospital rule to evaluate.

 

[The Chaz]

You could also use x=ln e^x and use the properties of the logarithm to combine the terms. The argument of the log will be an indeterminate form you can use the hospital rule to evaluate.

That's a pretty slick method!