Prove-that-lim-x-0-1-ax-1-b-1-x-a-b-

Question Number 44652 by rahul 19 last updated on 02/Oct/18

P r o v e t h a t \lim_{x \to 0} \frac{{(1 + a x)}^{\frac{1}{b}} - 1}{x} = \frac{a}{b} .

Commented by rahul 19 last updated on 02/Oct/18

w i t h o u t u s i n g L - h o s p i t a l r u l e .

Commented by maxmathsup by imad last updated on 02/Oct/18

w e h a v e {(1 + a x)}^{\frac{1}{b}} \sim 1 + \frac{a}{b} x (x \to 0) \Rightarrow

{(1 + a x)}^{\frac{1}{b}} - 1 \sim \frac{a}{b} x (x \to 0) \Rightarrow \frac{{(1 + a x)}^{\frac{1}{b}} - 1}{x} \sim \frac{a}{b} \Rightarrow

{l i m}_{x \to 0} \frac{{(1 + a x)}^{\frac{1}{b}} - 1}{x} = \frac{a}{b} .

Commented by rahul 19 last updated on 02/Oct/18

t h a n k s p r o f A b d o .

Commented by maxmathsup by imad last updated on 02/Oct/18

y o u a r e w e l c o m e s i r .

Answered by math1967 last updated on 02/Oct/18

\overset{l i m}{x} \to 0 \frac{a {{(1 + a x)}^{\frac{1}{b}} - 1}}{a x}

a \overset{l i m}{} x \to 0 \frac{{(1 + a x)}^{\frac{1}{b}} - 1}{a x}

\frac{a}{b} p r o v e d [\overset{L i m}{} x \to 0 \frac{{(1 + x)}^{n} - 1}{x} (n = a n y r a t i o n a l . .)

= n]

Commented by rahul 19 last updated on 02/Oct/18

t h a n k s s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

t = 1 + a x x \to 0 t \to 1

\lim_{t \to 1} \frac{a (t^{\frac{1}{b}} - 1)}{t - 1}

a \times \frac{1}{b} \times {(1)}^{\frac{1}{b} - 1}

= \frac{a}{b}

Commented by rahul 19 last updated on 02/Oct/18

t h a n k s s i r .