Limits ( 1 )

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

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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