Question

Calcule o limite no infinito

     $\Large\begin{matrix}\lim\limits_{x\to \infty}(2+3x^4)^{\frac{1}{1+2\ln x}}\end{matrix}$

sem recorrer às regras de L'Hôpital.

Answer 1

É dado o limite:

$L=\lim\limits_{x\to\infty} (2+3x^4)^{\dfrac{1}{1+2\ln(x)}}$

Vamos manipulá-lo:

$L=\lim\limits_{x\to\infty} (2+3x^4)^{\dfrac{1}{1+2\ln(x)}}\\\\ L=\lim\limits_{x\to\infty} e^{\ln\left((2+3x^4)^{\frac{1}{1+2\ln(x)}}\right)}\\\\ L=\lim\limits_{x\to\infty} e^{[\ln(2+3x^4)]/(1+2\ln(x))}}\\\\ L=e^{\lim\limits_{x\to\infty} [\ln(2+3x^4)]/(1+2\ln(x))}$

Seja L₂ o limite presente no expoente. Desenvolvendo-o:

$L_2=\lim\limits_{x\to\infty} \dfrac{\ln(2+3x^4)}{1+2\ln(x)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\ln(x^4(\frac{2}{x^4}+3))}{1+2\ln(x)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\ln(x^4)+\ln(\frac{2}{x^4}+3)}{1+2\ln(x)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{4\ln(x)+\ln(\frac{2}{x^4}+3)}{1+2\ln(x)}$

Dividindo o numerador e o denominador por $\ln(x)$:

$L_2=\lim\limits_{x\to\infty} \dfrac{4\ln(x)+\ln(\frac{2}{x^4}+3)}{1+2\ln(x)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\dfrac{4\ln(x)+\ln(\frac{2}{x^4}+3)}{\ln(x)}}{\dfrac{1+2\ln(x)}{\ln(x)}}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\dfrac{4\ln(x)}{\ln(x)}+\dfrac{\ln(\frac{2}{x^4}+3)}{\ln(x)}}{\dfrac{1}{\ln(x)}+\dfrac{2\ln(x)}{\ln(x)}}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{4+\dfrac{\ln(\frac{2}{x^4}+3)}{\ln(x)}}{\dfrac{1}{\ln(x)}+2}$

$L_2=\dfrac{4+0}{0+2}\\\\ L_2=\dfrac{4}{2}\\\\ L_2=2\\\\ \lim\limits_{x\to\infty} \dfrac{\ln(2+3x^4)}{1+2\ln(x)}=2$

Voltando ao limite L:

$L=e^{\lim\limits_{x\to\infty} [\ln(2+3x^4)]/(1+2\ln(x))}\\\\ L=e^2\\\\ \boxed{\lim\limits_{x\to\infty} (2+3x^4)^{\dfrac{1}{1+2\ln(x)}}=e^2}$