Limits and Derivatives Test - 21

Let $$\displaystyle S=\frac { lim }{ n\rightarrow \infty  } { \left[ \left( 1+\frac { 1 }{ { n }^{ 2 } }  \right) \left( 1+\frac { { 2 }^{ 2 } }{ { n }^{ 2 } }  \right) ...\left( 1+\frac { { n }^{ 2 } }{ { n }^{ 2 } }  \right)  \right]  }^{ \frac { 1 }{ n }  }$$

$$\displaystyle \Rightarrow logS=\frac { lim }{ n\rightarrow \infty  } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ log } \left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } }  \right) $$

$$\displaystyle =\int _{ 0 }^{ 1 }{ log } \left( 1+{ x }^{ 2 } \right) dx.=\left( xlog\left( 1+{ x }^{ 2 } \right)  \right) _{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \frac { { 2x }^{ 2 } }{ 1+{ x }^{ 2 } } dx } $$

$$\displaystyle \Rightarrow log\left( \frac { S }{ 2 }  \right) =2\left[ \int _{ 0 }^{ 1 }{ \frac { { dx } }{ 1+{ x }^{ 2 } }  } -\int _{ 0 }^{ 1 }{ dx }  \right] =2\left[ \frac { \pi  }{ 4 } -1 \right] =\frac { \pi -4 }{ 2 } $$

$$\displaystyle \Rightarrow S=2{ e }^{ \left( \frac { \pi -4 }{ 2 }  \right)  }$$

In the neighborhood of $$\:x=\dfrac{2\pi}3$$
$$\Rightarrow  y=-\cos\:x+\sin\:x$$
$$\Rightarrow \left ( \cfrac{dy}{dx} \right )=\sin\:x+\cos\:x$$
Thus at$$\:x=\dfrac{2\pi}3$$
$$\cfrac{dy}{dx}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}\left ( \sqrt{3}-1 \right )$$

$$\dfrac{1-\cos \left(x\right)}{\sin \left(x\right)}$$

$$\dfrac{d}{dx}\left(\tan^{-1} \left(\cot \left(x\right)\right)+\cot^{-1} \left(\tan \left(x\right)\right)\right)$$

Let,  $$\displaystyle y=\tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right )$$
$$\displaystyle y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}x$$
$$\therefore \displaystyle \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}$$

Give $$y = sec  x^0$$
$$= \displaystyle sec \frac{x \pi}{180^o} \left ( \because x^o = \displaystyle \frac{x \pi}{180^o} radians \right )$$
$$\therefore \displaystyle \frac{dy}{dx} = \frac{\pi}{180^o} sec  x^o  tan  x^o$$

Let $$x = \displaystyle \dfrac{1}{y}$$ then as $$x \rightarrow \infty \Rightarrow y \rightarrow 0$$
$$\therefore \displaystyle  \lim_{y \rightarrow 0} \left ( \sqrt{\dfrac{1}{y^2} + \dfrac{8}{y} + 3} - \sqrt{\dfrac{1}{y^2} + \dfrac{4}{y} + 3} \right )$$
$$\therefore \displaystyle \lim_{y \rightarrow 0}\frac{\sqrt{3y^2 + 8y + 1} - \sqrt{3y^2 + 4y + 1}}{y}$$
$$= \displaystyle \lim_{y \rightarrow 0} \left ( \dfrac{\displaystyle \dfrac{1}{2} (3y^2 + 8y + 1)^{-1/2} (6y + 8) - \dfrac{1}{2} (5y^2 + 4y + 1)^{-1/2} (6y + 4)}{1} \right )$$
$$= \displaystyle  \lim_{y \rightarrow 0} \left ( \frac{8}{2} - 2 \right ) = 2$$

$$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } }  } $$
LHL$$=f(0-0)=\displaystyle \lim _{ x\rightarrow 0 }{ (-h)\sin { { e }^{ (-1/h) } }  } $$
$$=-0\times \sin { \left( { e }^{ -\infty  } \right)  } $$
$$=-0\times \sin { (0)=0 } $$
$$=0\times$$ (a finite number between $$-1$$ to $$+1$$) $$\quad \left( \because -1\le \sin { x } \le 1 \right) $$
$$\because$$ LHL$$=$$RHL
$$\because$$ $$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } }  } $$ exist and equal to $$0$$

$$r=\sqrt { 2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi  }{ 4 }  \right)  } $$