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Limits and Continuity Test 1

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Limits and Continuity Test 1
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  • Question 1
    1 / -0
    Let $$f : R \to R$$ be a differentiable function satisfying $$f'(3) + f'(2) = 0$$.
    Then $$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$$ is equal to 
    Solution
    $$\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}$$  $$(1^{\infty}$$ form$$)$$

    $$\Rightarrow e^{\underset{x \to 0}{\lim} \dfrac{f(3+x)-f(2-x)-f(3)+f(2)}{x(1+f(2-x)-f(2))}}$$

    using L'Hopital
    $$\Rightarrow e^{\underset{x \to 0}{\lim}\dfrac{f'(3+x)+f'(2-x)}{-xf'(2-x)+(1+f(2-x)-f(2))}}$$

    $$\Rightarrow e^{\dfrac{f'(3) + f'(2)}{1}} = 1$$
  • Question 2
    1 / -0
    Let $$f$$ be a differentiable function such that $$f'(x) = 7- \dfrac{3}{4}\dfrac{f(x)}{x}, (x > 0)$$ and $$f(1) \neq 4$$.
    Then $$\underset{x\to 0^+}{\lim} xf \left(\dfrac{1}{x}\right) $$:
    Solution
    $$f'(x) = 7 -\dfrac{3}{4}\dfrac{f(x)}{x}$$    $$(x > 0)$$

    Given $$f(1) \neq 4$$    $$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = ?$$

    $$\dfrac{dy}{dx} + \dfrac{3}{4}\dfrac{y}{x} = 7$$  (This is LDE)

    IF $$=e^{\int \tfrac{3}{4x}dx} = e^{\tfrac{3}{4}ln|x|} = X^{\tfrac{3}{4}}$$

    $$y.x^{\frac{3}{4}} =\int 7.x^{\frac{3}{4}} dx$$

    $$y.x^{\frac{3}{4}} = 7.\dfrac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$$

    $$f(x) = 4x+C.x^{-\frac{3}{4}}$$

    $$f\left(\dfrac{1}{x}\right) = \dfrac{4}{x} + C.x^{\frac{3}{4}}$$

    $$\underset{x\to 0^+}{\lim} xf\left(\dfrac{1}{x}\right) = \underset{x\to 0^+}{\lim}\left(4+C.x^{\frac{7}{4}}\right) = 4$$
  • Question 3
    1 / -0
    If $$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$$, then $$\displaystyle \lim_{x\rightarrow 0} \dfrac {f'(x)}{x}$$.
    Solution
     $$f(x) = \begin{vmatrix} \cos x& x & 1\\ 2\sin x & x^{2} & 2x\ \\ \tan x & x & 1\end{vmatrix}$$
    $$=\cos x(x^{2}-2x^{2})-x(2\sin x-2x\tan x)+1(2x\sin x-x^{2}\tan x)$$
    $$=-x^{2}\cos x-2x\sin x+2x^{2}\tan x+2x\sin x-x^{2}\tan x$$
    $$=x^{2}\tan x-x^{2}\cos x$$
    $$=x^{2}(\tan x-\cos x)$$
    $$f'(x)=2x(\tan x-\cos x)+x^{2}(\sec^{2}x+\sin x)$$
    $$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=\lim_{x\rightarrow 0}\dfrac{2x(\tan x-\cos x)+x^{2}(\sec^2{x}+\sin x)}{x}$$
    $$=\displaystyle \lim_{x\rightarrow 0} 2(\tan x-\cos x)+x(\sec^{2}x+\sin x)$$
    $$=2(0-1)+0=-2$$
    Hence, $$\therefore \displaystyle \lim_{x\rightarrow 0} \dfrac{f'(x)}{x}=-2$$

  • Question 4
    1 / -0
    $$\lim _{ { x\rightarrow \pi /4 } } \dfrac { \cot ^{ { 3 } } x-\tan  x }{ \cos  (x+\pi /4) } $$  is
    Solution

  • Question 5
    1 / -0
    If the function $$f(x)= \left\{\begin{matrix}\dfrac {\sqrt {2+\cos x}-1}{(\pi-x)^2} & x\neq \pi \\ k & x=\pi \end{matrix}\right.$$ is continuous at $$x=\pi$$, then $$k$$ equals :
    Solution
    $$\displaystyle\lim _{ x\rightarrow \pi  }{ \frac { \sqrt { 2+\cos { x }  } -1 }{ (\pi -x)^{ 2 } }  } =\lim _{ x\rightarrow \pi  }{ \frac { 1+\cos { x }  }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x }  } +1) }  }   $$ [Rationalize numerator and denominator ]

                                        $$\displaystyle =\lim _{ x\rightarrow \pi  }{ \frac { 1-\cos { (\pi -x } ) }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x }  } +1) }  } $$

                                        $$=\displaystyle \lim _{ x\rightarrow \pi  }{ \frac { 2\sin ^{ 2 }{ \frac { (\pi -x) }{ 2 }  }  }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x }  } +1) }  } =\frac { 1 }{ 4 } $$

    So,$$ k =\dfrac{1}{4}$$                              .
  • Question 6
    1 / -0
    $$\underset{x\to 0}{\lim} \left(\dfrac{3x^2+2}{7x^2+2}\right)^{1/x^2}$$ is equal to:
    Solution
    $$L = \underset{x \to 0}{\lim} \left(\dfrac{3x^2 + 2}{7x^2 + 2}\right)^{\tfrac{1}{x^2}}$$           

    As it is a  $$1^{\infty}$$ form

    We know the formula,

    $$\Rightarrow$$$$\displaystyle\lim_{x \to 0} f(x)^{g(x)}= \displaystyle e^{\underset{x\to 0}{\lim} g(x) \left (f(x)-1\right)}$$

    $$= e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2}{7x^2 + 2}\displaystyle-1\right)}$$

    $$ = e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{3x^2+2-7x^2-2}{7x^2 + 2}\right)}$$

    $$ = e^{\underset{x\to 0}{\lim} \dfrac{1}{x^2} \left(\dfrac{-4x^2}{7x^2 + 2}\right)}$$

    $$=e^{\tfrac{-4}{2}}$$

    $$=e^{-2}$$

    $$=\dfrac1{e^{2}}$$

    $$\boxed{L=\dfrac1{e^{2}}}....Answer$$

    Hence option $$'A'$$ is the answer.
  • Question 7
    1 / -0
    If $$\lim _{ x\rightarrow \infty  }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } =A$$ and $$\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } =B$$, then which one of the following is correct?
    Solution
    As given $$A=\lim _{ x\rightarrow \infty  }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } =\lim _{ x\rightarrow \infty  }{ \cfrac { \sin { \left( \cfrac { 1 }{ x }  \right)  }  }{ \left( \cfrac { 1 }{ x }  \right)  }  } $$
    Let $$t=\cfrac { 1 }{ x } ;x\rightarrow \alpha ,t\rightarrow 0$$
    $$\Rightarrow A=\lim _{ t\rightarrow \infty  }{ \cfrac { \sin { t }  }{ t }  } =1\left[ \because \lim _{ t\rightarrow 0 }{ \cfrac { \sin { x }  }{ x } =1 }  \right] $$
    and $$B=\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x }  \right)  }  } \Rightarrow B=0\quad $$
    Therefore, $$A=1$$ and $$B=0$$ is correct
  • Question 8
    1 / -0
    $$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$$
    Solution
    $$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at} $$$$x\rightarrow 0$$
    $$but\ have\ finite\ value = 0$$
    $$0\times\ finite\ value\ =0$$
  • Question 9
    1 / -0
    If $$x$$ is very large, then $$\dfrac {2x}{1+x}$$ is
    Solution
    $$\dfrac {2x}{1+x}=\dfrac {2}{1+\dfrac {1}{x}}=\dfrac{2}{1+0}=2$$ 

    If  $$x\rightarrow \alpha$$ then $$\dfrac {1}{x}\rightarrow 0$$.
  • Question 10
    1 / -0
    $$\displaystyle \lim_{x\rightarrow \infty} \sin x$$ equals
    Solution
    Let $$L = \displaystyle \lim_{x\rightarrow \infty} \sin x$$
    Assume $$\displaystyle y = \frac{1}{x}$$ so as $$x\to \infty,  y \to 0$$
    $$\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}$$
    We know $$\sin x$$ lie between -1 to 1
    so let $$p =\sin x$$ as $$x\to \infty$$
    Thus left hand limit $$=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p$$
    and right hand limit $$=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p$$
    Clearly L.H.L $$\neq $$ R.H.L $$\Rightarrow$$ Required limit does not exist.
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