bemath-1-lim-x-2x-2-x-3-1-3-x-2-lim-x-1-1-x-1-sin-pix-3-0-x-2-f-t-dt-x-cos-pix-Find-f-4-

Question Number 111498 by bemath last updated on 04/Sep/20

\sqrt{bemath}

(1) \lim_{x \to \infty} \sqrt[3]{{2 x}^{2} - x^{3}} + x ?

(2) \lim_{x \to 1} {(\frac{1}{x})}^{\frac{1}{\sin π x}} ?

(3) \underset{0}{\int^{x^{2}}} f (t) dt = x \cos (π x) . Find f (4) .

Answered by john santu last updated on 04/Sep/20

l e t x = \frac{1}{t} \to {\begin{cases} x \to \infty \\ t \to 0 \end{cases}

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

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

L = \lim_{t \to 0} \frac{\sqrt[3]{2 t - 1} + 1}{t}

L = \lim_{t \to 0} \frac{2 t}{t [{(2 t - 1)}^{2} - (2 t - 1) + 1]}

L = \lim_{t \to 0} \frac{2}{{(2 t - 1)}^{2} + 2 - 2 t} = \frac{2}{3} .

Answered by bemath last updated on 04/Sep/20

Commented by bemath last updated on 04/Sep/20

t y p o i n \sqrt[3]{x} i t s h o u l d b e \sqrt[3]{x^{3}}

Answered by john santu last updated on 04/Sep/20

(2) \ln L = \lim_{x \to 1} (\frac{- \ln x}{\sin π x})

\ln L = \lim_{x \to 1} (\frac{- \frac{1}{x}}{π \cos π x})

\ln L = \frac{1}{π} \Rightarrow L = e^{\frac{1}{π}} = \sqrt[π]{e}

Answered by john santu last updated on 04/Sep/20

(3) \frac{d}{d x} [\int_{0}^{x^{2}} f (t) d t] = \frac{d}{d x} [x \cos π x]

\Rightarrow 2 x f (x^{2}) = \cos π x - π x \sin π x

p u t x = 2

\Rightarrow 4 f (4) = \cos 2 π - 2 π \sin 2 π

\Rightarrow f (4) = \frac{1}{4}

Answered by mathmax by abdo last updated on 04/Sep/20

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

= x - x {(1 - \frac{2}{x})}^{\frac{1}{3}} \sim x - x (1 - \frac{2}{3 x}) \Rightarrow f (x) \sim \frac{2}{3} \Rightarrow \lim_{x \to \infty} f (x) = \frac{2}{3}

Answered by mathmax by abdo last updated on 04/Sep/20

2) let f (x) = {(\frac{1}{x})}^{\frac{1}{\sin (π x)}} \Rightarrow f (x) = e^{\frac{- lnx}{\sin (π x)}} we have

\lim_{x \to 1} \frac{lnx}{\sin (π x)} =_{hospital} \lim_{x \to 1} \frac{1}{x (π \cos (π x)} = - \frac{1}{π} \Rightarrow

\lim_{x \to 1} f (x) = e^{\frac{1}{π}} =^{π} \sqrt{e}

Commented by bobhans last updated on 04/Sep/20

typo e^{- \frac{1}{π}} = \frac{1}{e^{\frac{1}{π}}}

Commented by mathmax by abdo last updated on 04/Sep/20

no typo!

Commented by bobhans last updated on 04/Sep/20

why ? you got e^{- \frac{1}{π}} ? it not same to

\sqrt[π]{e} ? ? ?

Answered by mathmax by abdo last updated on 04/Sep/20

3) we have \int_{0}^{x^{2}} f (t) dt = xcos (π x) let derivate \Rightarrow

2 xf (x^{2}) = \cos (π x) - π xsin (π x)

x = 2 we get 4 f (4) = \cos (2 π) - 2 π \sin (2 π) = 1 \Rightarrow f (4) = \frac{1}{4}

Answered by Dwaipayan Shikari last updated on 04/Sep/20

\lim_{x \to 1} {(\frac{1}{x})}^{\frac{1}{s i n π x}} = y

\frac{1}{s i n π x} l o g (\frac{1}{x}) = l o g y

\frac{\frac{1}{x} - 1}{s i n π x} \frac{l o g (1 + \frac{1}{x} - 1)}{(\frac{1}{x} - 1)} = l o g y

\lim_{w \to 0} \frac{\frac{1}{w + 1} - 1}{s i n (π w + π)} = l o g y

\lim_{w \to 0} \frac{\frac{- w}{w + 1}}{- s i n π w} = l o g y

\underset{x \to 0}{li m} \frac{\frac{- w}{w + 1}}{- w π} = l o g y

\frac{1}{π} = l o g y

y = e^{\frac{1}{π}}