How to prove $\lim_{x \to \infty} x^{(x+1)}-(x+1)^x = \infty$

Question

I try to prove this by using L'Hospital Rule but it doesn't work. I know it is infinity from wolframalpha but I don't know how to prove it.

Answer 1

Hint: $\ldots=x^x\left(x-\left(1+\frac1x\right)^x\right)\stackrel{x\to\infty}{\longrightarrow}\boxed{???}$

Answer 2

Use the estimate: $$n^{n+1}-(n+1)^n>n^n, n>3 \iff n^n(n-1)>(n+1)^n,n>3 \iff \\ n-1>\left(1+\frac 1n\right)^n, n>3.$$

Answer 3

$$\begin{align}\lim_{x \to \infty} x^{(x+1)}-(x+1)^x &=\lim_{x \to \infty} x^x\big(x-\big(1+\frac1x\big)^x\big)\\ &=\lim_{x \to \infty} x^x\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\tag{Product Rule}\\ &=\lim_{x \to \infty} e^{\ln{x^x}}\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\\ &=\lim_{x \to \infty} e^{x\ln{x}}\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\\ &=e^{\lim_{x \to \infty} x\lim_{x \to \infty}\ln{x}}\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\tag{Product rule}\\ &=e^{\infty\lim_{x \to \infty}\ln{x}}\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\\ &=\infty\lim_{x \to \infty} \big(x-\big(1+\frac1x\big)^x\big)\\ &=\infty \big(\lim_{x \to \infty} x-\lim_{x \to \infty}\big(1+\frac1x\big)^x\big)\\ &=\infty \big(\lim_{x \to \infty} x-\lim_{x \to \infty} e^{\ln{(1+\frac{1}{x})^x}}\big)\\ &=\infty \big(\lim_{x \to \infty} x- e^{\lim_{x \to \infty} x\ln{(1+\frac{1}{x})}}\big)\\ &=\infty \big(\lim_{x \to \infty} x- e^{\lim_{x \to \infty} \frac{\ln{(1+\frac{1}{x})}}{\frac{1}{x}}}\big)\\ &\stackrel{l'hop}{=}\infty \big(\lim_{x \to \infty} x- e^{\lim_{x \to \infty} {\frac{1}{1+\frac{1}{x}}}}\big)\\ &=\infty \big(\lim_{x \to \infty} x- e^{ {\frac{1}{\lim_{x \to \infty} 1+\frac{1}{x}}}}\big)\tag{Reciprocal Rule}\\ &=\infty \big(\lim_{x \to \infty} x- e^{ {\frac{1}{ 1+\frac{1}{\lim_{x \to \infty} x}}}}\big)\tag{Reciprocal Rule}\\ &=\infty \big(\lim_{x \to \infty} x- e^{ {\frac{1}{ 1+\frac{1}{\infty}}}}\big)\\ &=\infty \lim_{x \to \infty} x- e\\ &=\infty(\infty-e)\\ &=\infty\infty\\ &=\infty \end{align} $$