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Evaluate
$\underset{x\to \mathrm{\infty }}{lim}{x}^{2}\left({4}^{\frac{1}{x}}-{4}^{\frac{1}{1+x}}\right)$
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billyfcash5n
Write
$\frac{{4}^{t}-{4}^{t/\left(t+1\right)}}{{t}^{2}}={4}^{t/\left(t+1\right)}\cdot \frac{{4}^{{t}^{2}/\left(t+1\right)}-1}{\frac{{t}^{2}}{t+1}}\cdot \frac{1}{t+1}$
and use the standard limit $\underset{u\to 0}{lim}\frac{{a}^{u}-1}{u}=\mathrm{log}a$ to find
$\begin{array}{rl}\underset{t\to 0}{lim}\frac{{4}^{t}-{4}^{t/\left(t+1\right)}}{{t}^{2}}& =\underset{t\to 0}{lim}{4}^{t/\left(t+1\right)}\cdot \underset{t\to 0}{lim}\frac{{4}^{{t}^{2}/\left(t+1\right)}-1}{\frac{{t}^{2}}{t+1}}\cdot \underset{t\to 0}{lim}\frac{1}{t+1}\\ & =1\cdot \mathrm{log}4\cdot 1=\mathrm{log}4\end{array}$

2nalfq8
Mean Value Theorem, for $f\left(x\right)={4}^{x}$, with ${f}^{\prime }\left(x\right)={4}^{x}\mathrm{ln}4$: There exists a ${\xi }_{x}\in \left(1/x,1/\left(x+1\right)\right)$, such that
${4}^{\frac{1}{x}}-{4}^{\frac{1}{x+1}}=\left(\frac{1}{x}-\frac{1}{x+1}\right){4}^{{\xi }_{x}}\mathrm{ln}4=\frac{{4}^{{\xi }_{x}}\mathrm{ln}4}{x\left(x+1\right)}$
Clearly $\underset{x\to \mathrm{\infty }}{lim}{\xi }_{x}=0$, and hence $\underset{x\to \mathrm{\infty }}{lim}{4}^{{\xi }_{x}}=1$. Hence
$\underset{x\to \mathrm{\infty }}{lim}{x}^{2}\left({4}^{\frac{1}{x}}-{4}^{\frac{1}{x+1}}\right)=\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{2}}{x\left(x+1\right)}{4}^{{\xi }_{x}}\mathrm{ln}4=\mathrm{ln}4.$