Question

If $a,b,c$ are all real and non zero, then $\underset{x\to 0}{lim}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{1/x}$=

• A. $0$
• B. $(abc{)}^{1/3}$
• C. $(abc{)}^{-1/3}$
• D. $1$

Answer 1

The correct option is B $(abc{)}^{1/3}$

Let $y=\underset{x\to 0}{lim}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{\frac{1}{x}}$

Taking log on both the sides, we get

$\log y=\underset{x\to 0}{lim}\frac{\log (a^x+b^x+c^x)-\log 3}{x}$

Using L-Hospital rule, as form $\frac{0}{0}$

$\therefore \log y=\underset{x\to 0}{lim}\frac{a^x\log a+b^x\log b+c^x\log c}{a^x+b^x+c^x\cdot 1}$

{at x=0}

$=\frac{\log a+\log b+\log c}{3}$

$=\log (abc{)}^{\frac{1}{3}}$

$\therefore \log y=\log (abc{)}^{\frac{1}{3}}$

$y=(abc{)}^{\frac{1}{3}}$