Question

Evaluate: ${lim}_{x\to 1}\frac{\left(\ln \right(1+x)-\ln 2)({3.4}^{x-1}-3x)}{\left[\right(7+x{)}^{\frac{1}{3}}-(1+3x{)}^{\frac{1}{2}}]\cdot \sin (x-1)}$

Answer 1

$\underset{x\to 1}{lim}\frac{\left(ln\right(1+x)-ln2)(3.4^{x-1}-3x)}{\left[\right(7+x{)}^{\frac{1}{3}}-(1+3x{)}^{\frac{1}{2}}]\sin (x-1)}$

$\underset{h\to 0}{lim}\frac{ln\left(\frac{1+1+h}{2}\right)3[4^{h+1-1}-(1+h]}{\left[\right(7+1+h{)}^{\frac{1}{3}}-(1+3+3h{)}^{\frac{1}{2}}.\sin (1+h-1)}$

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

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

$\underset{h\to 0}{lim}\frac{\frac{3}{4}[\frac{4^h-1}{h}-1]}{\frac{1}{h}[\frac{h}{24}-\frac{3h}{8}]}$

$=-\frac{9}{4}ln\left(4/e\right)$.