# Find the solution $$\displaystyle\lim_{{{t}\to{0}}}{\frac{{{d}^{{{k}-{1}}}}}{{{\left.{d}{t}\right.}^{{{k}-{1}}}}}}{\left({\frac{{{t}\sqrt{{{1}+{t}}}}}{{{\log{\sqrt{{{1}+{t}}}}}}}}\right)}^{{k}}={2}{\left({k}+{2}\right)}^{{{k}-{1}}}$$

Justine White 2022-04-07 Answered
Find the solution
$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=2{\left(k+2\right)}^{k-1}$
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superpms01wks1
Complex analysis comes to the rescue. Using Cauchy differentiation formula:
${f}^{n-1}\left(0\right)=\frac{\left(n-1\right)!}{2\pi i}\oint \frac{f\left(z\right)}{{z}^{n}}dz$
Now
$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=\frac{\left(k-1\right)!}{2\pi i}\oint {\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}\frac{dt}{{t}^{k}}$
$\frac{\left(k-1\right)!}{2\pi i}\oint {\left(\frac{\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}dt$
Now performing the change of variable $t={e}^{u}-1$
$\underset{t\to 0}{lim}\frac{{d}^{k-1}}{{dt}^{k-1}}{\left(\frac{t\sqrt{1+t}}{\mathrm{log}\sqrt{1+t}}\right)}^{k}=\frac{\left(k-1\right)!}{2\pi i}\oint {\left(\frac{\mathrm{exp}\left(\frac{u}{2}\right)}{\frac{u}{2}}\right)}^{k}{e}^{u}du$
$={2}^{k}\left[\frac{\left(k-1\right)!}{2\pi i}\oint \frac{\mathrm{exp}\left(u\left(\frac{k}{2}+1\right)\right\}\left\{{u}^{k}\right\}du}{}$
$={2}^{k}\underset{u\to 0}{lim}\frac{{d}^{k-1}}{d{u}^{k-1}}{e}^{u\left(\frac{k}{2}+1\right)}={2}^{k}{\left(\frac{k}{2}+1\right)}^{k-1}=2{\left(k+2\right)}^{k-1}$