lim-x-0-x-2-1-1-5-x-1-1-3-x-1-1-3-x-1-

Question Number 102036 by Study last updated on 06/Jul/20

l i \underset{x \to 0}{m} \frac{\sqrt[5]{x^{2} - 1} + \sqrt[3]{x + 1}}{\sqrt[3]{x - 1} + \sqrt{x + 1}} = ?

Answered by john santu last updated on 06/Jul/20

L^{'} H o p i t a l

\lim_{x \to 0} \frac{\frac{2 x}{5 \sqrt[5]{{(x^{2} - 1)}^{4}}} + \frac{1}{3 \sqrt[3]{{(x + 1)}^{2}}}}{\frac{1}{3 \sqrt[3]{{(x - 1)}^{2}}} + \frac{1}{2 \sqrt{(x + 1)}}} =

\frac{\frac{1}{3}}{\frac{1}{3} + \frac{1}{2}} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}

(J S ⊛)

Answered by bemath last updated on 06/Jul/20

\lim_{x \to 0} \frac{\sqrt[3]{x + 1} - \sqrt[5]{1 - x^{2}}}{\sqrt{x + 1} - \sqrt[3]{1 - x}} =

\lim_{x \to 0} \frac{(\frac{x}{3} + 1) - (1 - \frac{x^{2}}{5})}{(\frac{x}{2} + 1) - (1 - \frac{x}{3})} =

\lim_{x \to 0} \frac{x (\frac{1}{3} + \frac{x}{5})}{x (\frac{1}{2} + \frac{1}{3})} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}

Commented by 1549442205 last updated on 06/Jul/20

{(1 + x)}^{m} = 1 + \frac{m}{1!} x + \frac{m (m - 1)}{2!} x^{2} + \frac{m (m - 1) (m - 2)}{3!} x^{3} + \dots

Commented by Study last updated on 06/Jul/20

w r i t e t h e f u r m o l l a

Commented by bemath last updated on 06/Jul/20

m a c l a u r i n s e r i e s

Commented by Dwaipayan Shikari last updated on 06/Jul/20

Answered by Dwaipayan Shikari last updated on 06/Jul/20

\lim_{x \to 0} \frac{\frac{2 x}{5 {(x^{2} - 1)}^{\frac{4}{5}}} + \frac{1}{3 {(x + 1)}^{\frac{2}{3}}}}{\frac{1}{3 {(x - 1)}^{\frac{2}{3}}} + \frac{1}{2 {(x + 1)}^{\frac{1}{2}}}} = \frac{\frac{2 x}{5} + \frac{1}{3}}{\frac{1}{3} + \frac{1}{2}} = \frac{2}{5}