Question

Problema nr.8
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Answer 1

$f(x) = \dfrac{3x^4}{x^2+1} \\ \\ l = \lim\limits_{x\to \infty}\Big(f(e^x)\Big)^{\frac{1}{x}}\\ \\ e^x = t \Rightarrow x = \ln t\\ x\to \infty \Rightarrow t\to \infty \\ \\ l = \lim\limits_{t\to \infty}\Big(f(t)\Big)^{\frac{1}{\ln t}} =\lim\limits_{t\to \infty}\Big(\dfrac{3t^4}{t^2+1}\Big)^{\frac{1}{\ln t}}}\\ \\ \ln (l) =\lim\limits_{t\to \infty}\ln \left[\Big(\dfrac{3t^4}{t^2+1}\Big)^{\frac{1}{\ln t}}}\right]=$

$=\lim\limits_{t\to \infty}\left[\dfrac{1}{\ln t}\ln \Big(\dfrac{3t^4}{t^2+1}\Big)\right]=\lim\limits_{t\to \infty}\dfrac{\ln\Big(\dfrac{3t^4}{t^2+1}\Big)}{\ln t}\overset{\frac{\infty}{\infty}}{=}\\ \\\overset{\frac{\infty}{\infty}}{=}\lim\limits_{t\to \infty}\dfrac{[\ln(3t^4)-\ln(t^2+1)]'}{(\ln t)'} = \lim\limits_{t\to \infty}\dfrac{\dfrac{12t^3}{3t^4}-\dfrac{2t}{t^2+1}}{\dfrac{1}{t}}= \\ \\ = \lim\limits_{t\to \infty }\Big(\dfrac{12t^4}{3t^4}-\dfrac{2t^2}{t^2+1}\Big) = \dfrac{12}{3}-2 = 4-2 = 2$

$\ln(l) = 2 \Rightarrow \boxed{l = e^2}$

Answer 2

Răspuns:

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