Limits and Continuity Test 16

$$\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( f\left( x \right)  \right)  } =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ g\left( \sin { x }  \right)  } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \left( \sin ^{ 2 }{ x } +1 \right)  } =0+1=1\\ \displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( f\left( x \right)  \right)  } =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ g\left( \sin { x }  \right)  } \\ =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \left( \sin ^{ 2 }{ x } +1 \right)  }=1$$

$$\because \tan\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right) = \dfrac{1 - \tan\dfrac{x}{2}}{1+\tan\dfrac{x}{2}}$$
$$\therefore \displaystyle \lim_{x\rightarrow \dfrac{\pi}{2}}\dfrac{\tan \left ( \dfrac{\pi }{4}-\dfrac{x}{2} \right )\left ( 1-\sin x \right )}{4\left ( \dfrac{\pi -2x}{4} \right )\left ( \pi -2x \right )^{2}}$$

Let $$\displaystyle P=\lim_{n\rightarrow \infty }\left ( \frac{n!}{n^{n}} \right )^{1/n}$$

Let  $$\sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+ \sqrt{(1 - \cos x)+...\infty)}}}=y$$

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

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

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

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

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

$$\displaystyle \underset { x\rightarrow 0 }{ lim } \frac { 1-cos^{ 3 }x+sin^{ 3 }x+\ell n(1+x^{ 3 })+\ell n(1+cos\, \, x) }{ x^{ 2 }-1+2\, cos^{ 2 }x+tan^{ 4 }x+sin^{ 3 }x } $$

$$\displaystyle =\frac { 1-1+0+\ln { \left( 1+0 \right)  } +\ln { \left( 1+1 \right)  }  }{ 0-1+2+0+0 } =\ln { 2 } $$

$$\displaystyle \lim _{ r\rightarrow \infty  }{ \dfrac { \cos { \dfrac { a }{ r } -\cos { \dfrac { b }{ r }  } \cos { \dfrac { c }{ r }  }  }  }{ \sin { \dfrac { b }{ r } \sin { \dfrac { c }{ r }  }  }  }  } $$

$$\displaystyle =\lim _{ r\rightarrow \infty  }{ \dfrac { \cos { \dfrac { a }{ r } -\dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b+c }{ r }  \right)  } +\cos { \left( \dfrac { b-c }{ r }  \right)  }  \right)  }  }{ \dfrac { 1 }{ 2 } \left( \cos { \left( \dfrac { b-c }{ r }  \right) -\cos { \left( \dfrac { b+c }{ r }  \right)  }  }  \right)  }  } $$

By Applying L Hospital,s Rule

$$\displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } }  } $$
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 $$0$$

Given $$\displaystyle \lim _{ x\rightarrow \infty  }{ \frac { { x }^{ 3 }+1 }{ { x }^{ 2 }+1 } -(ax+b)=2 } $$

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