How would you compute the limit $\lim_{x\to\infty} \left(\frac{1}{x^x+x}\right)^{1/x}$?

Question

I understand that you would use a natural log to break up because it is an indeterminate form, but that is what is seemingly giving me trouble.

Answer 1

We wish to evaluate the limit $ \lim_{x\to \infty}\frac{1}{\left(x^x+x\right)^{1/x}}$. We have

$$\frac{1}{\left(x^x+x\right)^{1/x}}=\frac{1}{x\left(1+\frac{1}{x^{x-1}}\right)^{1/x}}\tag 1$$

The term in parentheses approaches $1$ since $\frac1{x^{x-1}}\to 0$ and $1/x\to 0$ as $x\to \infty$.

Therefore, the limit of interest is $0$.

Answer 2

Note that

$$0< \left (\frac{1}{x^x+x}\right)^{1/x} < \left (\frac{1}{x^x}\right)^{1/x} = \frac{1}{x}.$$

Since $1/x \to 0,$ the squeeze theorem shows the limit of interest is $0.$

Answer 3

If you want to use logarithms, consider $$A=\left(\frac{1}{x^x+x}\right)^{1/x}\implies \log(A)=\frac1x\,\log\left(\frac{1}{x^x+x}\right)=-\frac1x\,\log\left({x^x+x}\right)$$ Now, write $$x^x+x=x^x\left(1+\frac 1{x^{x-1}}\right)$$ $$\log\left({x^x+x}\right)=\log(x^x)+\log\left(1+\frac 1{x^{x-1}}\right)=x\log(x)+\log\left(1+\frac 1{x^{x-1}}\right)$$ Since $x$ is large $$\log\left(1+\frac 1{x^{x-1}}\right)\sim \frac 1{x^{x-1}}$$ All of that make $$\log(A)\sim-\log(x)-\frac 1{x^{x}}\sim -\log(x)\implies A\sim \frac 1x$$ For $x=10$, using my pocket calculator $A=0.099999999990000000005$