Question

Arrange the following limits in the ascending order
a)$\underset{x\to 0}{lim}{\left(\frac{1^x+2^x+4^x}{3}\right)}^{\frac{1}{x}}$

b)$\underset{x\to 0}{lim}{\left(\frac{2^x+3^x+8^x+{27}^x}{4}\right)}^{\frac{1}{x}}$

c)$\underset{x\to 0}{lim}{\left[\frac{1^x+(\frac{1}{2}{)}^x+(\frac{1}{4}{)}^x}{3}\right]}^{\frac{1}{x}}$

d)$\underset{x\to 0}{lim}{\left[\frac{(\frac{1}{2}{)}^x+(\frac{1}{3}{)}^x+(\frac{1}{8}{)}^x+(\frac{1}{27}{)}^x}{4}\right]}^{\frac{1}{x}}$

• A. d,c,a,b
• B. d,a,b,c
• C. a,c,b,d
• D. a,b,c,d

Answer 1

The correct option is A d,c,a,b

We know that
$\underset{x\to 0}{lim}{\left(\frac{a_1^x+a_2^x+a_3^x+.....+a_n^x}{n}\right)}^{\frac{1}{n}}={(a_1{a}_2{a}_3.....a_n)}^{1/n}$
(a) $\underset{x\to 0}{lim}(\frac{1^x+2^x+4^x}{3}{)}^{\frac{1}{x}}$
$={\left(\mathrm{1.2.4}\right)}^{1/3}={\left(8\right)}^{1/3}=2$
b) ${lim}_{x\to 0}(\frac{2^x+3^x+8^x+{27}^x}{4}{)}^{\frac{1}{x}}$
$={\left(\mathrm{2.3.8.27}\right)}^{1/4}={\left(1296\right)}^{1/4}=6$
c)$\underset{x\to 0}{lim}[\frac{1^x+(\frac{1}{2}{)}^x+(\frac{1}{4}{)}^x}{3}{]}^{\frac{1}{x}}$
$={(1.\frac{1}{2}.\frac{1}{4})}^{1/3}={\left(\frac{1}{8}\right)}^{1/3}=\frac{1}{2}$
d)$\underset{x\to 0}{lim}[\frac{(\frac{1}{2}{)}^x+(\frac{1}{3}{)}^x+(\frac{1}{8}{)}^x+(\frac{1}{27}{)}^x}{4}{]}^{\frac{1}{x}}$

$={(\frac{1}{2}.\frac{1}{3}\frac{1}{8}\frac{1}{27})}^{1/4}={\left(\frac{1}{1296}\right)}^{1/4}=\frac{1}{6}$
Hence, ascending order of limits is $d,c,a,b$