Question-173765

Question Number 173765 by saly last updated on 17/Jul/22

Answered by cortano1 last updated on 18/Jul/22

L = \lim_{x \to 1} (\frac{3}{1 - \sqrt{x}} - \frac{2}{1 - \sqrt[3]{x}}) = ?

[l e t x = t^{6} \land t \to 1]

L = \lim_{t \to 1} (\frac{3}{1 - t^{3}} - \frac{2}{1 - t^{2}})

= \lim_{t \to 1} (\frac{3}{(1 - t) (1 + t + t^{2})} - \frac{2}{(1 - t) (1 + t)})

= \lim_{t \to 1} (\frac{3 t + 3 - 2 t^{2} - 2 t - 2}{(1 - t) (1 + t) (1 + t + t^{2})})

= \frac{1}{6} \times \lim_{t \to 1} (\frac{- (t - 1) (2 t + 1)}{- (t - 1)})

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

Commented by saly last updated on 18/Jul/22

T h a n k y o u

Answered by blackmamba last updated on 18/Jul/22

s e t x - 1 = r \land r \to 0

= \lim_{r \to 0} (\frac{3}{1 - \sqrt{1 + r}} - \frac{2}{1 - \sqrt[3]{1 + r}})

= \lim_{r \to 0} (\frac{3 (1 - \sqrt[3]{1 + r}) - 2 (1 - \sqrt{1 + r})}{(1 - \sqrt{1 + r}) (1 - \sqrt[3]{1 + r})})

= \lim_{r \to 0} (\frac{3 (1 - (1 + \frac{r}{3} - \frac{1}{9} r^{2})) - 2 (1 - (1 + \frac{1}{2} r - \frac{1}{8} r^{2}))}{(1 - (1 + \frac{1}{2} r)) (1 - (1 + \frac{1}{3} r))}

= \lim_{r \to 0} (\frac{3 (- \frac{r}{3} + \frac{r^{2}}{9}) - 2 (- \frac{r}{2} + \frac{r^{2}}{8})}{(- \frac{1}{2} r) (- \frac{1}{3} r)})

= \lim_{r \to 0} (\frac{\frac{r^{2}}{3} - \frac{r^{2}}{4}}{\frac{1}{6} r^{2}}) = 6 \times (\frac{4 - 3}{12}) = \frac{1}{2}

Answered by CElcedricjunior last updated on 18/Jul/22

- \frac{1}{2}

\dots . . l e c e l e b r e c e d r i c j u n i o r \dots \dots

Answered by Mathspace last updated on 18/Jul/22

f (x) = \frac{3}{1 - \sqrt{x}} - \frac{2}{1 -^{3} \sqrt{x}}

w e d o t h e c h a 7 g e m e n t x = 1 - t

(s o t \to 0) \Rightarrow

f (x) = f (1 - t) = \frac{3}{1 - \sqrt{1 - t}} - \frac{2}{1 -^{3} \sqrt{x}}

{(1 + u)}^{α} \sim 1 + α u + \frac{α (α - 1)}{2} u^{2}

{(1 - u)}^{α} \sim 1 - α u + \frac{α (α - 1)}{2} u^{2}

{(1 - t)}^{\frac{1}{2}} \sim 1 - \frac{t}{2} - \frac{1}{8} t^{2}

{(1 - t)}^{\frac{1}{3}} \sim 1 - \frac{t}{3} - \frac{2}{18} t^{2}

\Rightarrow f (1 - t) \sim \frac{3}{\frac{t}{2} + \frac{t^{2}}{8}} - \frac{2}{\frac{t}{3} + \frac{t^{2}}{9}}

= \frac{24}{4 t + t^{2}} - \frac{18}{3 t + t^{2}}

= \frac{24 (3 t + t^{2}) - 18 (4 t + t^{2})}{(4 t + t^{2}) (3 t + t^{2})}

= \frac{72 t + 24 t^{2} - 72 t - 18 t^{2}}{t^{2} (4 + t) (3 + t)}

= \frac{6}{(4 + t) (3 + t)} \to \frac{6}{12} = \frac{1}{2} (t \to 0)