Question-93695

Question Number 93695 by i jagooll last updated on 14/May/20

Commented by mathmax by abdo last updated on 14/May/20

f (x) = \frac{6 x -^{3} \sqrt{27 x^{3} + 2 x - 1}}{(^{3} \sqrt{8 x^{3} - x} + x)} \Rightarrow f (x) = \frac{6 x - 3 x {(1 + \frac{2}{27 x^{2}} - \frac{1}{27 x^{3}})}^{\frac{1}{3}}}{2 x {(1 - \frac{1}{8 x^{2}})}^{\frac{1}{3}} + x}

\Rightarrow f (x) \sim \frac{6 x - 3 x (1 + \frac{1}{3} (\frac{2}{27 x^{2}} - \frac{1}{27 x^{3}}))}{2 x (1 - \frac{1}{24 x^{2}}) + x} \sim \frac{3 x}{3 x} = 1 \Rightarrow {l i m}_{x \to - \infty} f (x) = 1

Answered by 1549442205 last updated on 14/May/20

\boldsymbol li \underset{\boldsymbol x \to - \infty}{\boldsymbol m} \frac{6 \boldsymbol x - \sqrt[3]{27 \boldsymbol x^{3} + 2 \boldsymbol x - 1}}{\sqrt[3]{8 \boldsymbol x^{3} - \boldsymbol x} + x} = \boldsymbol li \underset{\boldsymbol x \to - \infty}{\boldsymbol m} \frac{6 - \sqrt[3]{27 + \frac{2}{x^{2}} - \frac{1}{x^{3}}}}{\sqrt[3]{8 - \frac{1}{x^{2}}} + 1}

= \frac{6 - \sqrt[3]{27}}{\sqrt[3]{8} + 1} = \frac{6 - 3}{2 + 1} = 1

Commented by john santu last updated on 14/May/20

\lim_{x \to \infty} \neq \lim_{x \to - \infty} sir

Commented by john santu last updated on 14/May/20

your answer is wrong

Commented by 1549442205 last updated on 14/May/20

excuse me, I wrote fault

Answered by john santu last updated on 14/May/20

\lim_{x \to - \infty} (\frac{6 \boldsymbol x - \boldsymbol x \sqrt[3]{27 + \frac{2}{\boldsymbol x^{2}} - \frac{1}{\boldsymbol x^{3}}}}{\boldsymbol x \sqrt[3]{8 - \frac{1}{\boldsymbol x^{2}}} + \boldsymbol x}) =

\lim_{x \to - \infty} (\frac{- 6 + 3}{- 2 - 1}) = \frac{- 3}{- 3} = 1

Commented by i jagooll last updated on 14/May/20

cooll man ��

Commented by Ar Brandon last updated on 14/May/20

My man��