Question

Evaluate $\underset{x\to 1}{lim}\frac{\ell n\left[\frac{1+x}{2}\right]\times 3.(4^{x-1}-x)}{\left[\right(7+x{)}^{\frac{1}{3}}-(1+3x{)}^{\frac{1}{2}}]\sin (x-1)}$

• A. $\frac{9}{4}ln\frac{4}{e}$
• B. $-\frac{9}{4}ln\frac{4}{e}$
• C. $-\frac{3}{2}ln\frac{4}{e}$
• D. $-\frac{3}{2}ln\frac{2}{e}$

Answer 1

The correct option is B $-\frac{9}{4}ln\frac{4}{e}$

$\underset{x\to 1}{lim}\frac{\ell n\left[\frac{1+x}{2}\right]\times 3.(4^{x-1}-x)}{\left[\right(7+x{)}^{\frac{1}{3}}-(1+3x{)}^{\frac{1}{2}}]\sin (x-1)}$

Put x$\to$1+h

$=\underset{h\to 0}{lim}\frac{\ell n\left[\frac{2+h}{2}\right]\times 3(4^h-(1+h\left)\right)}{\left[\right(8+h{)}^{\frac{1}{3}}-(4+3h{)}^{1/2}].\sin h}$

$=\underset{h\to 0}{lim}\frac{\frac{\ell n[1+\frac{h}{2}]}{2\left(h/2\right)}.3.(4^h-(1+h\left)\right)}{2\left\{[1+\frac{h}{8}\times \frac{1}{3}+...]-[1+\frac{3h}{4}\times \frac{1}{2}+...]\right\}\frac{\sin h}{h}}$

$=\underset{h\to 0}{lim}\frac{\frac{3}{4}[\frac{4^h-1}{h}-1]}{\frac{1}{h}[\frac{h}{24}-\frac{3h}{8}]}=-\frac{9}{4}\ell n\frac{4}{e}.$