Limits ( 3 )

$\begin{aligned} L & = \lim_{n \to \infty} \left((n+1)\sqrt[n+1]{n+1} - n \sqrt[n] n\right) \\ & = \lim_{n \to \infty} \left(\blue{n\left(\sqrt[n+1]{n+1} - \sqrt[n] n \right)} + \sqrt[n+1]{n+1} \right) & \small \blue{\text{By theorem 1 (see reference)}} \\ & = \blue{\lim_{n \to \infty} n(n+1 - n) \frac {d\sqrt[n]n}{dn}} + \exp \left(\red{\lim_{n \to \infty}\frac {\ln(n+1)}{n+1}} \right) & \small \red{\text{An }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \blue{\lim_{n \to \infty} n\sqrt[n]n \left(\frac 1{n^2} - \frac {\ln n}{n^2} \right)} + \exp \left(\red{\lim_{n \to \infty} \frac {\frac 1{n+1}}1} \right) & \small \red{\text{Differentiate up and down w.r.t. }n} \\ & = \blue{\lim_{n \to \infty} \sqrt[n]n \left(\frac 1n - \frac {\ln n}n \right)} + e^\red 0 \\ & = \blue 0 + \red 1 = \boxed 1 \end{aligned}$

References:

• Theorem 1: $\displaystyle \lim_{n \to \infty} n(f(a_{n+1} - f(a_n)) = \lim_{n \to \infty} n(a_{n+1} - a_n) f'(a_n)$ . Source: New Methods For Calculations of Some Limits
• L'Hôpital's rule