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Limits and Derivatives Test - 23

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Limits and Derivatives Test - 23
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  • Question 1
    1 / -0
    The limit of $$\left\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\right\}$$ as $$x\rightarrow 0$$
    Solution
    Let $$p(x)=\cfrac { 1 }{ x } \sqrt { 1-x } ,q(x)=\sqrt { 1+\cfrac { 1 }{ { x }^{ 2 } }  } $$
    $$\lim _{ x\rightarrow { 0 }^{ + } }{ p\left( x \right)  } =\lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x } \sqrt { 1-x }  } =\infty \times \sqrt { 1-10 } $$
    $$=\infty $$
    $$\lim _{ x\rightarrow { 0 }^{ - } }{ p\left( x \right)  } =-\infty \times \sqrt { 1-0 } =-\infty $$
    as $$x\rightarrow { 0 }^{ + }\quad \cfrac { 1 }{ x } \rightarrow +\infty $$
    $$x\rightarrow { 0 }^{ - }\quad \cfrac { 1 }{ x } \rightarrow -\infty $$
    $$\Rightarrow p(x)$$ is discontinuous as $$x\rightarrow 0$$
    Let $$f\left( x \right) =p\left( x \right) -q\left( x \right) $$
    Since $$p\left( x \right) $$ is discontinuous $$f\left( x \right) $$ is also discontinuous by Algebra of continuous functions.
  • Question 2
    1 / -0
    If $$y = \tan^{-1} \left (\dfrac {1}{1 + x + x^{2}}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 3x + 2}\right ) + \tan^{-1} \left (\dfrac {1}{x^{2} + 5x + 6}\right ) + .... +$$ upto $$n$$ terms then $$\dfrac {dy}{dx}$$ at $$x = 0$$ and $$n = 1$$ is equal to
    Solution

    $$y= \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) + \tan^{-1}\left (\dfrac {1}{1 + (x + 1)(x + 2)}\right ) +\\ \quad\tan^{-1} \left (\dfrac {1}{1 + (x + 2)(x + 3)}\right ) + ..... + \tan^{-1} \left (\dfrac {1}{1 + (x + n - 1) (x + n)}\right )$$

    when $$n = 1$$
    $$y = \tan^{-1} \left (\dfrac {1}{1 + x(x + 1)}\right ) = \tan^{-1} \left (\dfrac {(x + 1) - x}{1 + x(x + 1)}\right ) = \tan^{-1} (x + 1) - \tan^{-1} x$$

    $$\Rightarrow \dfrac {dy}{dx} = \dfrac {1}{1 + (x + 1)^{2}} - \dfrac {1}{1 + x^{2}} \\ \therefore \dfrac {dy}{dx}]_{x = 0} = \dfrac {1}{1 + 1} - \dfrac {1}{1} = \dfrac {1}{2} - 1 = -\dfrac {1}{2}$$
  • Question 3
    1 / -0
    What is $$\displaystyle \lim_{x \rightarrow 0 }  x^2 \sin \left(\frac{1}{x}\right)$$ equal to ? 
    Solution
    $$\displaystyle \lim _{ x\rightarrow 0 }{ { x }^{ 2 }\sin { \left( \cfrac { 1 }{ x }  \right)  }  } $$
    $$={ 0 }^{ 2 }\times \sin { \left( \cfrac { 1 }{ 0 }  \right)  } \quad \left[ -1\le \sin { \left( \cfrac { 1 }{ x }  \right)  } \le 1 \right] $$
    $$=0$$
  • Question 4
    1 / -0
    If $$f\left( x \right) =\begin{vmatrix} \sin { x }  & \cos { x }  & \tan { x }  \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}$$, then $$\displaystyle\lim _{ x\rightarrow 0 }{ \dfrac { f\left( x \right)  }{ { x }^{ 2 } }  } $$ is
    Solution
    $$f(x)=\begin{vmatrix} sinx & cosx & tanx \\ { x }^{ 3 } & { x }^{ 2 } & x \\ 2x & 1 & 1 \end{vmatrix}\\ \Longrightarrow f(x)=sinx({ x }^{ 2 }-x)-cosx({ x }^{ 3 }-2{ x }^{ 2 })+tanx({ x }^{ 3 }-2{ x }^{ 3 })\\ \Longrightarrow f(x)={ x }^{ 2 }sinx-xsinx-{ x }^{ 3 }cosx+2{ x }^{ 2 }cosx-{ x }^{ 3 }tanx\\ \Longrightarrow \dfrac { f(x) }{ { x }^{ 2 } } =sinx-xcosx+2cosx-xtanx-\dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =\underset { x\longrightarrow 0 }{ lim } sinx-\underset { x\longrightarrow 0 }{ lim } xcosx+\underset { x\longrightarrow 0 }{ lim } 2cosx-\underset { x\longrightarrow 0 }{ lim } xtanx-\underset { x\longrightarrow 0 }{ lim } \dfrac { sinx }{ x } \\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =0-0+2-0-1\\ \Longrightarrow \underset { x\longrightarrow 0 }{ lim } \dfrac { f(x) }{ { x }^{ 2 } } =1\\ \\ $$
  • Question 5
    1 / -0
    $$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$$ is equal to :
    Solution
    $$\displaystyle\lim_{x\rightarrow \frac{\pi}{2}}(\pi - 2x^{\cos x})$$ 
    $$=\pi - 2\left(\dfrac{\pi}{2}\right)^{\cos \frac{\pi}{2}}$$
    $$=\pi-2\left(\dfrac{\pi}{2}\right)^{0}$$
    $$=\pi-2$$
  • Question 6
    1 / -0
    The derivative of $$ln(x+\sin x)$$ with respect to $$(x+\cos x)$$ is
    Solution
    Let $$u=ln(x+\sin x)$$   and   $$v=x+\cos x$$
    Now,
    $$\cfrac{du}{dx}=\cfrac1{x+\sin x}(1+\cos x)$$ .... $$(i)$$
    and $$\cfrac{dv}{dx}=1-\sin x$$ ..... $$(ii)$$
    Now, we can find derivative of $$u$$ with respect to $$v$$ by dividing $$(i)$$ by $$(ii)$$, we get
    $$\cfrac{du/dx}{dv/dx}=\cfrac{(1+\cos x)}{(x+\sin x)(1-\sin x)}$$
    or, $$\cfrac{du}{dv}=\cfrac{1+\cos x}{(x+\sin x)(1-\sin x)}$$
    Hence, A is the correct option.
  • Question 7
    1 / -0
    Let   $$C(\theta)=\displaystyle\sum _{n=0}^{\infty}\dfrac{\cos(n\theta)}{n!}$$
    Which of the following statements is FALSE? 
    Solution
    $$C(0)=e$$
    $$C(\pi)=\dfrac{1}{e}$$
    Hence, option A and B are correct

    $$C'(\theta)= -\displaystyle \sum_{n=0}^{\infty }\dfrac{\sin\theta }{(n-1)!}$$
    $$C'(\theta)=0$$
    Therefore, option D is false because $$C'(\theta) =0$$ for $$n=0$$
  • Question 8
    1 / -0
    $$\displaystyle\lim_{x\rightarrow\frac{\pi}{6}}\frac{\sin\left(x-\displaystyle\frac{\pi}{6}\right)}{\sqrt{3-2cos x}}$$ is equal to :
    Solution
    $$ \lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \dfrac { \sin { (x-\cfrac { \pi  }{ 6 }  } ) }{ \sqrt { 3-2\cos { x }  }  }  } $$
    at  $$\cfrac { \pi }{ 6 }$$  we get the numerator as $$\sin { (x-\cfrac { \pi  }{ 6 } ) } = 0 $$ 
    and the denominator as $$  \sqrt { 3-2\cos { x }  } =\sqrt { 3-\sqrt { 3 }  } $$   
    So,it is not an indeterminant form as we have $$\dfrac{0}{\sqrt{3 - \sqrt{3}} }$$
    Hence the answer will be zero. 
  • Question 9
    1 / -0
    $$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x}$$ is equal to
    Solution
    $$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} = \displaystyle \lim_{h\rightarrow 0}\dfrac {\tan \left (\dfrac {\pi}{4} + h\right ) - 1}{\cos 2\left (\dfrac {\pi}{4} + h\right )}$$
    $$\left [\because x = \dfrac {\pi}{4} + h\right ]$$

    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {\left (\dfrac {1 + \tan h}{1 - \tan h}\right ) - 1}{\cos \left (\dfrac {\pi}{2} + 2h\right )}$$

    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {1 + \tan h - 1 + \tan h}{-\sin 2h (1 - \tan h)}$$
    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-2\tan h}{2\sin h \cos h (1 - \tan h)}$$
    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-1}{\cos^{2} h (1 - \tan h)} = -1$$
    Hence, option D is correct.
  • Question 10
    1 / -0
    The value of $$\displaystyle \lim _{ x\rightarrow \pi /6 }{ \cfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 }  } $$ is
    Solution

    $$L=\lim _{ x\rightarrow { \pi  }/{ 6 } }{ \dfrac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } -1 }  } \\ L=\dfrac { 2\sin ^{ 2 }{ \left( \dfrac { \pi  }{ 6 }  \right)  } +\sin { \left( \dfrac { \pi  }{ 6 }  \right)  } -1 }{ 2\sin ^{ 2 }{ \left( \dfrac { \pi  }{ 6 }  \right)  } -3\sin { \left( \dfrac { \pi  }{ 6 }  \right)  } -1 } \\ L=\dfrac { 2.\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 2 } -1 }{ 2.\dfrac { 1 }{ 4 } -3.\dfrac { 1 }{ 2 } -1 } =\dfrac { 0 }{ -2 }= 0 $$

    So option $$D$$ is correct

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