Question

$\underset{x\to 0}{lim}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{2/x},a,b,c>0$ is equal to

• A. $0$
• B. $(abc{)}^{2/3}$
• C. $abc$
• D. $(abc{)}^{1/2}$

Answer 1

Answer: (B)

$(abc{)}^{2/3}$

$\underset{x\to 0}{lim}(\frac{a^x+b^x+c^x}{3}{)}^{\frac{2}{x}}$

The above is of the form :$1^{\infty }$

$=e^{{lim}_{x\to 0}\left(\frac{2}{x}\right)(\frac{a^x+b^2+c^x}{3}-1)}$

$=e^{{lim}_{x\to 0}\left(\frac{2}{x}\right)\left(\frac{a^x-1+b^x-1+c^x-1}{3}\right)}$

$=e^{{lim}_{x\to 0}\left(\frac{2}{3}\right)[\frac{a^{x-1}}{x}+\frac{b^x-1}{x}+\frac{c^x-1}{x}]}$

$=e^{\frac{2}{3}(lna+lnb+lnc)}\{\therefore \underset{x\to 0}{lim}\frac{t^x-1}{x}=ln(t\left)\right\}$

$=e^{\frac{2}{3}\left[ln\right(abc\left)\right]}$ $=e^{ln\left(\right(abc{)}^{2/3})}$

$=(abc{)}^{2/3}$