lim-x-0-1-x-a-1-x-for-a-R-Don-t-using-L-hospital-rules-

Question Number 25379 by A1B1C1D1 last updated on 09/Dec/17

\lim_{x \to 0} (\frac{{(1 + x)}^{a} - 1}{x}) for a \in R

{Don}^{'} t using L^{'} hospital rules .

Answered by Rasheed.Sindhi last updated on 10/Dec/17

\lim_{x \to 0} (\frac{{(1 + x)}^{n} - 1}{x})

\lim_{x \to 0} (\frac{(1 + \frac{n}{1} x + \frac{n (n - 1)}{1.2} x^{2} + \frac{n (n - 1) (n - 2)}{1.2.3} x^{3} + \dots - 1}{x})

\lim_{x \to 0} (\frac{(\frac{n}{1} x + \frac{n (n - 1)}{1.2} x^{2} + \frac{n (n - 1) (n - 2)}{1.2.3} x^{3} + \dots)}{x})

\lim_{x \to 0} (\frac{n}{1} + \frac{n (n - 1)}{1.2} x + \frac{n (n - 1) (n - 2)}{1.2.3} x^{2} + \dots)

= n

Commented by A1B1C1D1 last updated on 09/Dec/17

Thank you

Answered by moxhix last updated on 10/Dec/17

a n o t h e r w a y

p u t f (x) = {(1 + x)}^{a}

f^{'} (x) = a {(1 + x)}^{a - 1} \to f^{'} (0) = a

f^{'} (0) = \underset{x \to 0}{l i m} \frac{{(1 + x)}^{a} - {(1 + 0)}^{a}}{x - 0}

∴ \underset{x \to 0}{l i m} \frac{{(1 + x)}^{a} - 1}{x} = a

Facebook Tweet Pin