Question

The value of $\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{log\sqrt{5}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\infty )}$ is?

• A. $2$
• B. $4$
• C. $8$
• D. $0$

Answer 1

Answer: (B)

$4$

$\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{\log \sqrt{5}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty )}$

$=\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{\log \sqrt{5}\left(\frac{1}{4}\frac{1}{1-\frac{1}{2}}\right)}$

because $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....\infty$ (infinite GP sum)

$=\frac{a}{1-r}=\frac{1}{4}\frac{1}{1-\frac{1}{2}}$

$=\frac{1}{2}$

$\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{{\log }_{5^{\frac{1}{2}}}\frac{1}{2}}=\underset{n\to \infty }{lim}{\left(\frac{1}{5}\right)}^{2{\log }_5\frac{1}{2}}$ ........... Using ${\log }_{b^n}a=\frac{1}{n}{\log }_{b}a$

$=\underset{n\to \infty }{lim}\frac{1}{5^{{\log }_5\frac{1}{4}}}$ .......... Using $r{\log }_{b}a={\log }_b{a}^r$

$=\underset{n\to \infty }{lim}\frac{1}{\frac{1}{4}}=4$