Encontrar el límite siguiente: $L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$

Question

Encuentra el siguiente límite:

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Creo que usar la expansión de Taylor da$$L=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$ o hay una solución alternativa, pero no sé.

Answer 1

Deje$b_n=\sqrt[n]{n}-1=n^{1/n}-1$ y$a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}$ para que$$a_n=\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\frac{\left(2\sqrt[n]{n}-2+1\right)^n}{n^2}=\frac{1}{n^2}\operatorname{e}^{n\ln(1+2b_n)}$ $ Para$n\to \infty, \;b_n\to 0$ porque$b_n=\operatorname{e}^{\frac{\ln n}{n}}-1$ y para$n\to \infty$,$\frac{\ln n}{n}\to 0$. Entonces para$n\to\infty$ tenemos$\ln(1+2b_n)\sim 2b_n$, y$b_n=n^{1/n}-1=\operatorname{e}^{\frac{\ln n}{n}}-1\sim 1+\frac{\ln n}{n}-1=\frac{\ln n}{n}$ para que $$ n \ ln (1 +2b_n) \ sim n2b_n \ sim n2 \ frac {\ ln n} {n} = \ n n n ^ 2. $$ Finalmente para$n\to\infty$ $$ a_n = \ frac {1} {n ^ 2} \ operatorname {e} ^ {n \ ln (1 +2b_n)} \ sim \ frac {1} {n ^ 2 } \ operatorname {e} ^ {\ ln n ^ 2} = \ frac {n ^ 2} {n ^ 2} = 1 $$ que es $$ \ lim_ {n \ to \ infty} \ frac {\ left ( 2 \ sqrt [n] {n} -1 \ right) ^ n} {n ^ 2} = 1 $$

Answer 2

$$ \begin{align} L&=\lim_{n\to \infty}\frac{\left(2\sqrt[n]{n}-1\right)^n}{n^2}=\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n}-1}{n^{2/n}}\right)^n \\ &= \operatorname{e}^{\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n}-1}{n^{2/n}}-1\right)n}=\operatorname{e}^{-\lim_{n\to \infty}\left(\frac{n^{1/n}-1}{n^{1/n}}\right)^2\cdot n}\\ &= \operatorname{e}^{-\lim_{n\to \infty}\left(\frac{\operatorname{e}^{\frac{\ln n}{n}}-1}{{\frac{\ln n}{n}}}\right)^2\cdot\left(\frac{\ln n}{n}\right)^2\cdot \frac{1}{n^{2/n}}\cdot n}\\ &=\operatorname{e}^{-1\cdot 1\cdot \lim_{n\to \infty}\frac{\ln^2n}{n}}=\operatorname{e}^{0}= 1 \end {align} $$