Question

The value of $\underset{x\to 1}{lim}\frac{\left(\ln \right(1+x)-\ln 2)(3\cdot 4^{x-1}-3x)}{\left(\right(7+x{)}^{1/3}-(1+3x{)}^{1/2})\sin \pi x}$ is

• A. $\frac{2}{3\pi }(\ln 4-\frac{1}{2})$
• B. $\frac{4}{3\pi }(\ln 4-\frac{2}{3})$
• C. $\frac{9}{4\pi }(\ln 4-1)$
• D. $\frac{4}{3\pi }(\ln 2-\frac{2}{3})$

Answer 1

The correct option is C $\frac{9}{4\pi }(\ln 4-1)$

We have,
$L=\underset{x\to 1}{lim}\frac{\left(\ln \right(1+x)-\ln 2)(3\cdot 4^{x-1}-3x)}{\left(\right(7+x{)}^{1/3}-(1+3x{)}^{1/2})\sin \pi x}$
Let $x=1+h$
As $x\to 1,h\to 0$

$L=\underset{h\to 0}{lim}\frac{\left(\ln \right(2+h)-\ln 2)(3\cdot 4^h-3-3h)}{\left(\right(8+h{)}^{1/3}-(4+3h{)}^{1/2})(-\sin \pi h)}$

$=\underset{h\to 0}{lim}\frac{3\left(\frac{\ln (1+\frac{h}{2})}{h}\right)(\frac{4^h-1}{h}-1)}{(\frac{(8+h{)}^{1/3}-8^{1/3}}{h}-\frac{3\left(\right(4+3h{)}^{1/2}-4^{1/2})}{3h})(-\pi \frac{\sin \pi h}{\pi h})}$

$=\frac{3\times \frac{1}{2}\times (\ln 4-1)}{(\frac{1}{3}\times \frac{1}{4}-3\times \frac{1}{2}\times \frac{1}{2})(-\pi )}$

$=\frac{\frac{3}{2}(\ln 4-1)}{\frac{2\pi }{3}}$

$=\frac{9}{4\pi }(\ln 4-1)$