lim-a-x-1-b-x-1-c-x-1-a-b-c-1-x-x-0-

Question Number 127358 by Fareed last updated on 29/Dec/20

\lim {(\frac{a^{x + 1} + b^{x + 1} + c^{x + 1}}{a + b + c})}^{\frac{1}{x}} = ?

x \Rightarrow 0

Answered by Ar Brandon last updated on 29/Dec/20

Ψ = \lim_{x \to 0} {(\frac{a^{x + 1} + b^{x + 1} + c^{x + 1}}{a + b + c})}^{\frac{1}{x}}

\ln Ψ = \lim_{x \to 0} \frac{1}{x} \ln (\frac{a^{x + 1} + b^{x + 1} + c^{x + 1}}{a + b + c}) = \frac{1}{0} \ln (\frac{a + b + c}{a + b + c}) = \frac{0}{0}

= \lim_{x \to 0} {\frac{a^{x + 1} lna + b^{x + 1} lnb + c^{x + 1} lnc}{a + b + c} \cdot \frac{a + b + c}{a^{x + 1} + b^{x + 1} + c^{x + 1}}}

= \frac{alna + blnb + clnc}{a + b + c}

Ψ = e^{\frac{alna + blnb + clnc}{a + b + c}}

Commented by Study last updated on 29/Dec/20

w h e a r i s t h e \frac{1}{x} d i r i v a t i o n ? ? ?

Commented by Study last updated on 29/Dec/20

? ? ? ? ? ?

Commented by Ar Brandon last updated on 29/Dec/20

\lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{0}{0} \Rightarrow \lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{f^{'} (x)}{g^{'} (x)}

\lim_{x \to 0} \frac{1}{x} \ln (\frac{a^{x + 1} + b^{x + 1} + c^{x + 1}}{a + b + c}) = \lim_{x \to 0} \frac{\ln^{'} (\frac{a^{x + 1} + b^{x + 1} + c^{x + 1}}{a + b + c})}{x^{'}}