Calculate limit of $\sqrt[n]{2^n - n}$ Calculate limit of $\sqrt[n]{2^n - n}$.
I know that lim $\sqrt[n]{2^n - n} \le 2$, but don't know where to go from here.

The exponential function is continuous, so $$\lim_{n\rightarrow \infty}\sqrt[n]{2^n -n} = \lim_{n\rightarrow \infty} e^{\frac{\ln(2^n-n)}{n}} = e^{(\lim_{n\rightarrow \infty}\frac{\ln(2^n-n)}{n})}$$
Then you could use l'Hospital to show that $$\lim_{n\rightarrow \infty}\frac{\ln(2^n-n)}{n} = \ln(2)$$
So then you'd have $$\lim_{n\rightarrow \infty}\sqrt[n]{2^n -n} = e^{\ln(2)} = 2$$
