Proving Limits: How to Show x^x Approaches Infinity as x Approaches Infinity?

[FallArk]

1.
\(\displaystyle \log_{e}\left({x}\right)\to-\infty\) as \(\displaystyle x\to{0}^{+}\)
2.
\(\displaystyle {x}^{x}\to\infty\) as \(\displaystyle x\to\infty\)

I know how to prove that \(\displaystyle \log_{e}\left({x}\right)\) approaches \(\displaystyle \infty\) as x approaches \(\displaystyle \infty\) by using the definition given in the book, not sure how to use that to prove the first problem.

 

[Euge]

Hi FallArk,

In the first problem, set $y = 1/x$. Then $y \to \infty$ as $x \to 0+$, and $\log_e x = -\log_e y$. So it suffices to compute $\lim\limits_{y\to \infty} -\log y$. For the second problem, compute the limit of $\ln(x^x) = x\ln x$ first.

 

[FallArk]

Hi FallArk,

In the first problem, set $y = 1/x$. Then $y \to \infty$ as $x \to 0+$, and $\log_e x = -\log_e y$. So it suffices to compute $\lim\limits_{y\to \infty} -\log y$. For the second problem, compute the limit of $\ln(x^x) = x\ln x$ first.

I'm still a bit confused about the second problem. is it using the idea that \(\displaystyle {x}^{x}\) > xlnx and by limit comparison \(\displaystyle {x}^{x} \implies \infty\)

 

[I like Serena]

I'm still a bit confused about the second problem. is it using the idea that \(\displaystyle {x}^{x}\) > xlnx and by limit comparison \(\displaystyle {x}^{x} \implies \infty\)

Perhaps we can simplify it a bit.
For $x>1$ we have that $x^x >x$.
So if $x\to\infty$ so does $x^x\to\infty$.

 

[Euge]

I'm still a bit confused about the second problem. is it using the idea that \(\displaystyle {x}^{x}\) > xlnx and by limit comparison \(\displaystyle {x}^{x} \implies \infty\)

That idea works (your inequality is valid for $x > 1$), but that's not what I meant. What I meant is the following. Suppose $L = \lim\limits_{x\to \infty} x^x$. Then $\ln L = \lim\limits_{x\to \infty} \ln(x^x) = \lim\limits_{x\to \infty} x\ln x$. Prove $\lim\limits_{x\to \infty} x\ln x = \infty$. Then $L = \infty$.