Q:

Finding the limit of $2^{-1/\sqrt {n}}$ as $n \rightarrow \infty$.

Finding the limit of $2^{-1/\sqrt {n}}$ as $n \rightarrow \infty$. I feel like I can put $n = m^2$, them my limit will be $2^{-1/m}$ as $n \rightarrow \infty$. which is equal to lim $\frac{1}{2^{1/m}}$, then how can I solve this last limit?
Could anyone help me please?

A:

Since $2^x$ is continuous, $\displaystyle{\lim_{n \to \infty}} 2 ^{-\frac{1}{\sqrt{n}}}  = 2^{\displaystyle{\lim_{n \to \infty}} -\frac{1}{\sqrt{n}}}= 2^0 = 1$

A:

It's easy to get more intuition in problems like this. If the limit exists, let it be $L$, then
$$
\ln L = \ln \left( \lim_{n \to \infty} 2^{-1/\sqrt{n}}\right)
      = \lim_{n \to \infty} \ln\left( 2^{-1/\sqrt{n}}\right)
      = \lim_{n \to \infty} \ln 2 \cdot \frac{-1}{\sqrt{n}}
      = 0
$$
so $L = e^{\ln L} = e^0 = 1$.

