Limit in a sequence

Question

I know that $\lim\sqrt[n]{a}=1$(where $a > 0$ is a real number).

I know also that $\lim{\frac{1}{n}}=0$.

But, can you explain me why $\lim\sqrt[n]{2 + \frac{1}{n}}= 1$ ?

Answer 1

Note that $\sqrt[n]{2+\frac{1}{n}}$ is Squeezed betweeen $1$ and $\sqrt[n]{3}$.

Answer 2

Since $2<2+\frac1n\le 3$, yuo can compare your sequence with $\sqrt[n]2$ and $\sqrt[n]3$.

Answer 3

More brute and straightforward: take log of the function, use L'Hospital's rule: $$ \frac{\log (2+\frac{1}{x})}{x} \to-\frac{1}{x(x+2)} $$ This limit tends to 0 as $x \to \infty$. Now if you exponentiate you get $e^0=1$