Question

$\underset{x\to 1}{lim}\frac{(\log (1+x)-\log 2)(3.4^{x-1}-3x)}{\left\{{(7+x)}_3^1\right)-{(1+3x)}^{\frac{1}{2}}\left)\right\}\sin \pi x}$

Answer 1

$L={lim}_{x\to 1}\frac{(\log (1+x)-\log 2)({3.4}^{x-1}-3x)}{({(7+x)}^{\frac{1}{3}}-{(1+3x)}^{\frac{1}{2}})\sin \pi x}L=\frac{{lim}_{x\to 1}(\log (1+x)-\log 2).{lim}_{x\to 1}(\frac{3}{4}.4^x-3x)}{{lim}_{x\to 1}[{(7+x)}^{\frac{1}{3}}-{(1+3x)}^{\frac{1}{2}}]{lim}_{x\to 1}\sin \pi x}$

Applying L'Hospital's rule

$L=\frac{{lim}_{x\to 1}\left(\frac{1}{x+1}\right)\times {lim}_{x\to 0}(\frac{3}{4}.4^x\log 4-3)}{{lim}_{x\to 1}\pi \cos \pi x\times {lim}_{x\to 1}[\frac{1}{3}{(7+x)}^{\frac{-2}{3}}-\frac{3}{2}{(1+3x)}^{\frac{-1}{2}}]}=\frac{\left(\frac{1}{1+1}\right)\times (\frac{3}{4}\times 4^0\times \log 4-3)}{\pi \cos 0\times [\frac{1}{3}{\left(8\right)}^{\frac{-2}{3}}-\frac{3}{2}{\left(4\right)}^{\frac{-1}{2}}]}=\frac{\frac{1}{2}\times (\frac{3}{4}\log 4-3)}{\pi \times [\frac{1}{3}\times \frac{1}{4}-\frac{3}{2}\times \frac{1}{2}]}=\frac{\frac{3}{2}(\frac{\log 4}{4}-1)}{\pi \times (\frac{1}{12}-\frac{3}{4})}=\frac{3}{2}\times \frac{(\frac{\log 4}{4}-1)}{\pi \times \left(\frac{1-9}{12}\right)}=\frac{3\times 12}{2\times \pi }(\frac{\log 4}{4}-1)\times \frac{1}{-8}=\frac{-9}{4\pi }(\frac{\log 4}{4}-1)$