Ta có: \(\sqrt[n]{a}=a^{\dfrac{1}{n}}\)
\(\Rightarrow lim_{n\rightarrow\infty}\sqrt[n]{a}=lim_{n\rightarrow\infty}a^{\dfrac{1}{n}}=1\)
\(\Rightarrow lim_{n\rightarrow0^+}\sqrt[n]{a}=lim_{n\rightarrow0^+}a^{\dfrac{1}{n}}=+\infty\)
\(\Rightarrow lim_{n\rightarrow0^-}\sqrt[n]{a}=lim_{n\rightarrow0^-}a^{\dfrac{1}{n}}=0\)