The limit of the expression $\lim_{n\rightarrow \infty }(2-2^{\frac{1}{n+1}})^{n}$ evaluates to $\frac{1}{2}$. The solution involves rewriting the limit as $L=e^{\lim_{n\rightarrow \infty }(n)(1-2^{\frac{1}{n+1}})$, leading to the logarithmic form $\ln(L)=\lim_{n\to\infty}\left(\frac{\ln\left(2-2^{\frac{1}{n+1}}\right)}{\dfrac{1}{n}}\right)$. Applying L'Hôpital's rule resolves the indeterminate form, confirming that $L=\frac{1}{2}$. The discussion also references the remarkable limit $\lim_{x\rightarrow 0}(1+x)^{\frac{1}{x}}=e$ as a potential method for solving similar limits.
PREREQUISITESStudents and professionals in mathematics, particularly those studying calculus and limits, as well as educators looking for examples of limit evaluations and applications of L'Hôpital's rule.