lim-x-4-2-x-x-1-3-2-8-x-x-

Question Number 66640 by hmamarques1994@gmail.com last updated on 18/Aug/19

\underset{x \to 4}{l i m} \frac{\sqrt[3]{2 + x \sqrt{x}} - 2}{8 - x \sqrt{x}} = ?

Commented by kaivan.ahmadi last updated on 18/Aug/19

x \to 4^{+} : \frac{\sqrt[3]{10} - 2}{8 - 8^{+}} = \frac{\sqrt[3]{10} - 2}{0^{-}} = - \infty

x \to 4^{-} : \frac{\sqrt[3]{10} - 2}{0^{+}} = + \infty

Commented by kaivan.ahmadi last updated on 18/Aug/19

i f y o u m e a n {l i m}_{x \to 4} \frac{\sqrt[3]{4 + 2 \sqrt{x}} - 2}{8 - x \sqrt{x}} t h e n i t i s e q u a l t o

{l i m}_{x \to 4} \frac{\frac{1}{3 \sqrt{x} \sqrt[3]{{(4 + 2 \sqrt{x})}^{2}}}}{- \sqrt{x} - \frac{\sqrt{x}}{2}} = \frac{\frac{1}{3 \times 2 \times 4}}{- 2 - 1} = - \frac{1}{72}

Commented by mathmax by abdo last updated on 18/Aug/19

p e r h a p s t h e Q i s f i n d {l i m}_{x \to 4} \frac{(^{3} \sqrt{x \sqrt{x}} - 2}{8 - x \sqrt{x}} o r {l i m}_{x \to 4} \frac{(^{3} \sqrt{4 + 2 \sqrt{x}} - 2}{8 - x \sqrt{x}}

l e t f i n d {l i m}_{x \to 4} \frac{(^{3} \sqrt{x \sqrt{x}} - 2}{8 - x \sqrt{x}} = {l i m}_{x \to 4} A (x) c h a n g e m e n t

(^{3} \sqrt{x \sqrt{x}}) = t g i v e x \sqrt{x} = t^{3} \Rightarrow {l i m}_{x \to 4} A (x) = {l i m}_{t \to 2} \frac{t - 2}{8 - t^{3}}

= - {l i m}_{t \to 2} \frac{t - 2}{(t - 2) (t^{2} + 2 t + 4)} = {l i m}_{t \to 2} \frac{- 1}{t^{2} + 2 t + 4} = \frac{- 1}{12}

l e t f i n d {l i m}_{x \to 4} \frac{(^{3} \sqrt{4 + 2 \sqrt{x}} - 2}{8 - x \sqrt{x}} = {l i m}_{x \to 4} B (x) w e d l t h e c h a n g e m e n t

(^{3} \sqrt{4 + 2 \sqrt{x}}) = t \Rightarrow 4 + 2 \sqrt{x} = t^{3} \Rightarrow 2 \sqrt{x} = t^{3} - 4 \Rightarrow \sqrt{x} = \frac{t^{3}}{2} - 2 \Rightarrow

x = {(\frac{t^{3}}{2} - 2)}^{2} \Rightarrow {l i m}_{x \to 4} B (x) = {l i m}_{t \to 2} \frac{t - 2}{8 - {(\frac{t^{3}}{2} - 2)}^{3}}

= {l i m}_{t \to 2} \frac{t - 2}{8 - {(\frac{t^{3} - 4}{2})}^{3}} = {l i m}_{t \to 2} 8 \times \frac{t - 2}{64 - {(t^{3} - 4)}^{3}}

= {l i m}_{t \to 2} \frac{8 (t - 2)}{64 - (t^{9} - 3 t^{6} (4) + 3 t^{3} 4^{2} - 4^{3})}

= {l i m}_{t \to 2} \frac{8 (t - 2)}{64 - t^{9} + 12 t^{6} - 48 t^{3} + 64} = {l i m}_{t \to 2} \frac{8 (t - 2)}{- t^{9} + 12 t^{6} - 48 t^{3} + 128}

=_{h o p} {l i m}_{t \to 2} \frac{8}{- 9 t^{8} + 12 \times 6 t^{5} - 48 \times 3 t^{2}} = \frac{8}{- 9 . 2^{8} + 12 \times 6 . 2^{5} - 48.3 . 2^{2}}

= \dots . .

Commented by hmamarques1994@gmail.com last updated on 18/Aug/19

G o o d!

Commented by Sayantan chakraborty last updated on 18/Aug/19

ANSWER ? ? ? ?

Commented by mathmax by abdo last updated on 18/Aug/19

c o m p l e t e t h e c a l c u l u s s i r .