Question

Calcule o limite no infinito

     $\Large\begin{matrix}\lim\limits_{x\to \infty} (25x^3+4x-1)^{\frac{1}{\ln(x^2+7x-5)}}\end{matrix}$

sem recorrer às regras de L'Hôpital.

Answer 1

É dado o seguinte limite:

$L=\lim\limits_{x\to\infty} (25x^3+4x-1)^{\dfrac{1}{\ln(x^2+7x-5)}}$

Manipulando-o:

$L=\lim\limits_{x\to\infty} (25x^3+4x-1)^{\dfrac{1}{\ln(x^2+7x-5)}}\\\\ L=\lim\limits_{x\to\infty} e^{\ln((25x^3+4x-1)^{\frac{1}{\ln(x^2+7x-5)}})}\\\\ L=\lim\limits_{x\to\infty} e^{[\ln((25x^3+4x-1)]/[\ln(x^2+7x-5)]}\\\\ L=e^{\lim\limits_{x\to\infty} [\ln((25x^3+4x-1)]/[\ln(x^2+7x-5)]}$

Seja L₂ o limite presente no expoente da expressão anterior. Vamos calculá-lo:

$L_2=\lim\limits_{x\to\infty} \dfrac{\ln(25x^3+4x-1)}{\ln(x^2+7x-5)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\ln\left(x^3\left(25+\dfrac{4}{x^2}-\dfrac{1}{x^3}\right)\right)}{\ln\left(x^2\left(1+\dfrac{7}{x}-\dfrac{5}{x^2}\right)\right)}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{\ln\left(x^3\right)+\ln\left(25+\dfrac{4}{x^2}-\dfrac{1}{x^3}\right)}{\ln\left(x^2\right)+\ln\left(1+\dfrac{7}{x}-\dfrac{5}{x^2}\right)}$

$L_2=\lim\limits_{x\to\infty} \dfrac{3\ln(x)+\ln\left(25+\dfrac{4}{x^2}-\dfrac{1}{x^3}\right)}{2\ln(x)+\ln\left(1+\dfrac{7}{x}-\dfrac{5}{x^2}\right)}$

Dividindo o numerador e o denominador por ln(x), obtemos:

$L_2=\lim\limits_{x\to\infty} \dfrac{\dfrac{1}{\ln(x)}\left[3\ln(x)+\ln\left(25+\dfrac{4}{x^2}-\dfrac{1}{x^3}\right)\right]}{\dfrac{1}{\ln(x)}\left[2\ln(x)+\ln\left(1+\dfrac{7}{x}-\dfrac{5}{x^2}\right)\right]}\\\\ L_2=\lim\limits_{x\to\infty} \dfrac{3+\dfrac{1}{\ln(x)}\cdot\ln\left(25+\dfrac{4}{x^2}-\dfrac{1}{x^3}\right)}{2+\dfrac{1}{\ln(x)}\cdot\ln\left(1+\dfrac{7}{x}-\dfrac{5}{x^2}\right)}\\\\ L_2=\dfrac{3+0}{2+0}\\\\ \boxed{L_2=\dfrac{3}{2}}$

Agora basta substituir na última expressão obtida de L:

$L=e^{\lim\limits_{x\to\infty} [\ln((25x^3+4x-1)]/[\ln(x^2+7x-5)]}=e^{L_2}\\\\ L=e^{\frac{3}{2}}\\\\ \boxed{\boxed{\lim\limits_{x\to\infty} (25x^3+4x-1)^{1/\ln(x^2+7x-5)}=e^{\frac{3}{2}}}}$