L'Hospital for $ \infty- \infty$

Question

I want to find out the value of this expression.

$\lim_{x\to\infty}(\frac{1}{2}\ln^2(x)+x\ln\pi)-\lim_{x\to0}(\frac{1}{2}\ln^2(x)+x\ln\pi)$

By plugging in the values, I get $\infty-\infty$. According to wolframalpha, the value is undefined.

By just seeing that it's $\infty-\infty$, can we then conclude that it's undefined? Won't L'hospitale or other tricks work here? How to really know whether it's undefined?

EDIT:

Thanks for all the answers. Actually I was trying to find whether the $\int_{0}^\infty\frac{\ln(x\pi^x)dx}{x}$ converges or not. And after integrating I got the above-mentioned results. So I was lost there.

EDIT2: Sorry for messing this question up. I asked another question concerning the integral here:Improper Integration $\int_{0}^\infty\frac{\ln(x\pi^x)\,dx}{x}$ converges or not.

Thanks alot.

Answer 1

We have that

$$\lim_{x\to\infty} \frac{1}{2}\ln^2(x)+x\ln\pi =\infty$$

$$\lim_{x\to 0} \frac{1}{2}\ln^2(x)+x\ln\pi =\infty$$

then we can't evaluate

$$\lim_{x\to\infty} \frac{1}{2}\ln^2(x)+x\ln\pi-\lim_{x\to\infty} \frac{1}{2}\ln^2(x)+x\ln\pi$$

since $\infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.

Otherwise for finite limit we have

$$\lim_{x\to a} f(x) =L_1 \land \lim_{x\to b} g(x) =L_2 \implies \lim_{x\to a} f(x) \pm \lim_{x\to b} g(x) =L_1 \pm L_2$$

What we can evaluate is

$$\lim_{x\to\infty}\left[ \left(\frac{1}{2}\ln^2(x)+x\ln\pi\right) - \left(\frac{1}{2}\ln^2(1/x)+(1/x)\ln\pi\right)\right]$$

Answer 2

In writing $$\lim_{x\to\infty}\left(\frac{1}{2}\ln^2(x)+x\ln\pi\right)-\lim_{x\to0}\left(\frac{1}{2}\ln^2(x)+x\ln\pi\right),$$ you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).

Answer 3

An indeterminate form $\infty - \infty$ might be something like $\lim_{x \to +\infty} (a(x)-b(x))$. But here you have two different limits: $\lim_{x \to \infty} a(x) - \lim_{t \to 0} b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.

Answer 4

There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $\lim_{x\to a} f(x)=\lim_{x\to b} g(x)=\infty$.

When you ask what is $\lim_{x\to a}f(x)-\lim_{x\to b}g(x)$ you are asking what is the value of $\infty-\infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.

If you had $a=b$, then you could ask whether $\lim_{x\to a}(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$\infty-\infty$". But that is very different: in here, the $\infty$ symbol is only a potential infinity, and the expression "$\infty-\infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).

But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.