calculate limit

Question

$\lim_{x\rightarrow 0}\left ( \frac{a^{x}+b^{x}}{2} \right )^{^{\frac{1}{x}}}$

Answer 1

Answer = ab

Let $y = \lim_{ x \to 0} \left({\frac {a^x+b^x}{2}}\right)^{\frac{1}{x}}$

taking log on both sides

$\log y = \lim_{ x \to 0}\log \left( {\frac {a^x+b^x}{2}}\right)^{\frac{1}{x}}$

$= \lim_{ x \to 0} \frac{\log \frac {a^x+b^x}{2}} {x}$

This is of the form $\frac{0}{0}$ and so applying L'Hôpital's rule

$\log y = \lim_{x \to 0} \frac {1}{a^x+b^x} . \left(a^x\log a+b^x \log b \right)$

$=\frac{1}{2} . \left( \log a+ \log b \right)$

$=\log \sqrt {ab}$

So, $y= \sqrt{ab}$.

Comments

@shreshtha check it and correct it

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i made a calculation mistake here.. logy=limx→0   2 / a^x+b^x.(a^xloga+b^xlogb)

it should be-

logy=limx→0    2 / a^x+b^x ( 1/2a^xloga+1/2b^xlogb)

logy=limx→0      2 / a^x+b^x  1/2.(a^xloga+b^xlogb)

log y= 1/2(log a + log b)

logy = 1/2 log ab

log y= log√ab

y=√ab 

thanks again :) :)

Answer 2

$\lim_{x\rightarrow 0}((a^{x}+b^{x})/2)^{1/x}$   gives $1^{\infty }$    form so we apply standard form

$e^{\lim_{x\rightarrow 0}((f(x)-1)*(g(x)))}$       

here in this question f(x) =$((a^{x}+b^{x})/2)$      and   g(x)=1/x

so we apply standard form $e^{\lim_{x\rightarrow 0}((a^{x}+b^{x})/2)-1)*1/x}$ 

now we calculate  for$lim_{x\rightarrow 0}((a^{x}+b^{x})/2)-1)*1/x$ which is 0/0 form now we apply l hospital rule that gives

(a^x lna+b^xlnb)/2  =1/2*(lnab)=ln (ab)^1/2

$\therefore e^{ln_{e}\sqrt{ab}}=\sqrt{ab}$

Answer 3

I'm getting (ab)^(1/2)

Comments

yes its answer is (ab)^1/2