¿Podemos resolver este límite sin exprimir el teorema?

Question

Me preguntaba si podemos resolver este límite sin usar el teorema de squeeze (sándwich). ps

Answer 1

Ciertamente. $$ \ lim_ {n \ to + \ infty} \ exp \ left (\ ln \ left ((3 ^ n + 5 ^ n) ^ {2 / n} \ right) \ right) = \ lim_ {n \ to + \ infty} \ exp \ left (\ frac {2 \ ln (3 ^ n + 5 ^ n)} {n} \ right) = \ exp \ left (\ lim_ {n \ to + \ infty} \ frac { 2 \ ln (3 ^ n + 5 ^ n)} {n} \ right) $$

Ahora, resolvamos$\displaystyle \lim_{n \to +\infty}\frac{2\ln(3^n + 5^n)}{n}$. Tenemos$\displaystyle \frac{+\infty}{+\infty}$, así que podemos aplicar la regla de l'Hospital directamente, dándonos: $$ \ lim_ {n \ to + \ infty} \ frac {2 \ ln (3) 3 ^ n + 2 \ ln (5) 5 ^ n} {3 ^ n + 5 ^ n} = 2 \ ln (5) $$

Entonces, el límite es$\displaystyle \exp(2 \ln(5)) = 25$.

Answer 2

ps

Ahora usemos una regla de L'Hopital:

ps

Answer 3

$$ \begin{align} &\lim_{n\to \infty}(3^n+5^n)^{2/n}\\ =&\lim_{n\to \infty}\exp\left( {2\log(3^n+5^n)/n}\right)\\ =&\exp\left({\lim_{n\to \infty}{2\log(3^n+5^n)/n}}\right)\\ =&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)3^n+2\log(5)5^n}{3^n+5^n}}}\right)\\ =&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)(3/5)^n+2\log(5)}{(3/5)^n+1}}}\right)\\ =&\exp\left({{\frac{[0]+2\log(5)}{[0]+1}}}\right)\\ =&\exp\left(2\log 5\right)\\ =&25\\ \end {align} $$

Answer 4

Puede escribir, ya que$n$ tiende a$+\infty$,

$$ \begin{align} (3^n+5^n)^{2/n}&=e^{\frac2n\log \left(3^n+5^n \right)}\\\\ &=e^{\frac2n\log \left(5^n\right)+ \frac2n\log \left(1+(3/5)^n \right)}\\\\ &=e^{2\log 5+ \frac2n \log \left(1+(3/5)^n \right)}\\\\ &=e^{2\log 5+ \frac2n (3/5)^n }\\\\ &\sim e^{2\log 5}\times e^0\\\\ &\sim25 \end {align} $$ donde hemos usado $$ \ log (1 + x) \ sim_0 x$$ with $ x: = (3/5) ^ n$, $ n $ siendo genial

Answer 5

$$\lim{n \to \infty }\sqrt[n]{(3^n+5^n)^2}=\\lim{n \to \infty }\sqrt[n]{(5^n((\frac{3}{5})^n+1))^2}=\5^2\lim{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=\$$as we now $$\lim{n \to \infty }(\frac{3}{5})^n=0$$so $$5^2\lim_{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=5^2*\sqrt[n]{(0+1)^2}=25$$