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Limits and Continuity Test 32

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Limits and Continuity Test 32
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  • Question 1
    1 / -0
    If function $$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix}  &  \end{matrix} \end{cases}$$
    is continuous at $$x=2$$ then value of $$m$$ will be:
    Solution
    $$f(x)=\begin{cases} \begin{matrix} \dfrac{\sin 3x}{x}; & x \ne 0 \end{matrix} \\ \begin{matrix} m; & x=0 \end{matrix} \\ \begin{matrix} & \end{matrix} \end{cases}$$
    $$f(0)=m$$
    $$f(0+0)=\displaystyle \lim_{h \rightarrow 0}f(0+h)$$
    $$\displaystyle \lim_{h \rightarrow 0} \dfrac{\sin 3 (0+h)}{0+h}$$
    $$\displaystyle \lim_{h \rightarrow 0}3 \dfrac{\sin 3 h}{3h}$$
    $$=3 \times 1$$
    $$=3$$
    At $$x=a$$ function is continuous.
    So, $$f(0)=f(0+0)$$
    $$m=3$$
    Hence, option $$(a)$$ is correct.
  • Question 2
    1 / -0
    The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$$ is
    Solution
    $$\displaystyle \lim_{x \rightarrow 0} \dfrac {xe^{x} - \log_{e} (1 + x)}{x^{2}}$$
    $$x \left [1 + x + \dfrac {x^{2}}{2!} + \dfrac {x^{3}}{3!} + ....\right ]$$
    $$= \displaystyle \lim_{x \rightarrow 0} \dfrac {-\left [x - \dfrac {x^{2}}{2} + \dfrac {x^{3}}{3} - \dfrac {x^{4}}{4} + ....\right ]}{x^{2}}$$
    $$= \displaystyle \lim_{x \rightarrow 0} \dfrac {\left [x + x^{2} + \dfrac {x^{3}}{2!} + \dfrac {x^{4}}{3!} + .... - x + \dfrac {x^{2}}{2} - \dfrac {x^{3}}{3} + \dfrac {x^{4}}{4} - ....\right ]}{x^{2}}$$
    $$= \displaystyle \lim_{x \rightarrow 0} \dfrac {x^{2} \left [1 + \dfrac {x}{2!} + \dfrac {x^{2}}{3!} + ... + \dfrac {1}{2} - \dfrac {x}{3} + \dfrac {x^{2}}{4} - .... \right ]}{x^{2}}$$
    $$= \left [1 + 0 + 0 + ... + \dfrac {1}{2} - 0 + 0 ... \right ] = 1 + \dfrac {1}{2} = \dfrac {3}{2}$$
    Hence, option (D) is correct.
  • Question 3
    1 / -0
    The value of $$\displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin x}{x}$$ is
    Solution
    $$\displaystyle \lim_{x \rightarrow \infty} \dfrac {\sin x}{x}$$
    Let $$x = 1/y$$ or $$y = \dfrac {1}{x}$$
    Som $$x\rightarrow \infty \Rightarrow y \rightarrow 0$$
    $$\therefore \displaystyle \lim_{x \rightarrow \infty} \left (\dfrac {\sin x}{x}\right ) = \displaystyle \lim_{y \rightarrow 0} y \sin \left (\dfrac {1}{y}\right )$$
    $$= \displaystyle \lim_{y \rightarrow 0} y \cdot \displaystyle \lim_{y \rightarrow 0} \sin \dfrac {1}{y}$$
    $$= 0\times \text {(any variable quantity)}$$
    $$= 0$$
    Hence, option (A) is correct.
  • Question 4
    1 / -0
    The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$ is
    Solution
    $$\displaystyle \lim_{x\rightarrow 0} \dfrac {1 - \cos x}{x^{2}}$$
    $$= \displaystyle \lim_{x\rightarrow 0} \dfrac {1 - 1 + 2\sin^{2} x/2}{x^{2}}$$
    $$= \displaystyle \lim_{x\rightarrow 0} \dfrac {2\sin^{2} x/2}{x^{2}}$$
    $$= \displaystyle \lim_{x\rightarrow 0} 2\left (\dfrac {\sin x/2}{x/2}\right )^{2} \times \dfrac {1}{4}$$
    $$= \dfrac {1}{2}\times \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin x/2}{x/2}\right )^{2}$$
    $$= \dfrac {1}{2} \times 1 = \dfrac {1}{2}$$
    Hence option (B) is correct.
  • Question 5
    1 / -0
    The value of $$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$$ is
    Solution
    $$\displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{\tan x}\right )^{4}$$
    $$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\times 3x\right )^{4}\times \dfrac {1}{\tan^{4} x}$$
    $$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {3^{4}}{\dfrac {\tan^{4}x}{x^{4}}}$$
    $$= \displaystyle \lim_{x\rightarrow 0} \left (\dfrac {\sin 3x}{3x}\right )^{4} \times \dfrac {1}{\left (\dfrac {\tan x}{x}\right )^{4}} \times 81$$
    $$= 1\times 1\times 81 = 81$$
    Hence, option (B) is correct.
  • Question 6
    1 / -0
    The value of $$\displaystyle \lim_{x\rightarrow \infty} \sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}$$ is
    Solution
    $$\displaystyle \lim_{\rightarrow \infty} x \sin \dfrac {\pi}{4x}\cos \dfrac {\pi}{4x}$$
    $$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {2\sin \dfrac {\pi}{4x} \cos \dfrac {\pi}{4x}}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {1}{2} \times \dfrac {\pi}{2x}$$
    $$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (2\times \dfrac {\pi}{4x}\right )}{\left (\dfrac {\pi}{2x}\right )} \times \dfrac {\pi}{4}$$
    $$= \dfrac {\pi}{4}\times \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (\dfrac {\pi}{2x}\right )}{\left (\dfrac {\pi}{2x}\right )} = \dfrac {\pi}{4}\times 1 = \dfrac {\pi}{4}$$
    Hence, option (A) is correct.
  • Question 7
    1 / -0
    Let $$\alpha(a)$$ and $$\beta(a)$$ be the roots of the equation $$(\sqrt[3]{1+a}-1)x^{2}+(\sqrt{1+a}-1){x}+(\sqrt[6]{1+a}-1)=0$$ where $$a>-1$$. Then $$ \underset{a\rightarrow 0^{+}}{\lim}\alpha(a)$$ and $$ \underset{a\rightarrow 0^{+}}{\lim}\beta(a)$$ are 
    Solution
    Let $$1+a=y$$. 

    The equation becomes $$(y^{\frac {1}{3}}-1)x^2+(y^{\frac {1}{2}}-1)x+(y^{\frac {1}{6}}-1)=0$$

    Dividing by $$y-1$$, we get

    $$\displaystyle \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$$

    When $$a\rightarrow 0, y\rightarrow 1$$.

    Hence, taking $$\underset {y\rightarrow 1}{\lim}$$ on both the sides,
    $$\Rightarrow \displaystyle \underset {y\rightarrow 1}{\lim} \left(\dfrac {y^{\frac {1}{3}}-1}{y-1}\right)x^2+\left(\dfrac {y^{\frac {1}{2}}-1}{y-1}\right)x+\left(\dfrac {y^{\frac {1}{6}}-1}{y-1}\right)=0$$

    $$\Rightarrow \dfrac {1}{3}x^2+\cfrac {1}{2}x+\cfrac {1}{6}=0$$             ...[Using $$\underset {x\rightarrow a}{\lim}\cfrac {x^n-a^n}{x-a}=na^{n-1}]$$

    $$\Rightarrow 2x^2+3x+1=0$$

    Solving the quadratic equation, we get

    $$\Rightarrow x=-1$$ or $$x=\dfrac {-1}{2}$$.

    Since $$\alpha (a)$$ and $$\beta (a)$$ are roots of the given equation, we get

    $$\underset {a\rightarrow 0^+}{\lim}\alpha (a)=-1$$ and $$\underset {a\rightarrow 0^+}{\lim}\beta (a)=\dfrac {-1}{2}$$.
  • Question 8
    1 / -0
    Let $$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$$. If $$f$$ is continuous at $$x = 0$$, then the value of $$a / b$$ is:
    Solution
    $$f(x)=\displaystyle \frac{(256+ax)^{1/8}-2}{(32+bx)^{1/5}-2}$$
    Given , $$f$$ is continuous at $$x=0$$.
    $$ \displaystyle \lim _{ x\rightarrow 0 } f(x)=f(0)$$
    Now, $$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ \frac { (256+ax)^{ 1/8 }-2 }{ (32+bx)^{ 1/5 }-2 }  } $$
    It is of the form $$\displaystyle \frac{0}{0}$$, so applying L-Hospital's rule
    $$\displaystyle =\lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ 8 } a(256+ax)^{ -7/8 } }{ \frac { 1 }{ 5 } b(32+bx)^{ -4/5 } }  } $$
    $$\displaystyle = \frac { 5a }{ 64b } $$
    Hence,$$\displaystyle \frac { 5a }{ 64b }=f(0)$$
    $$\Rightarrow \displaystyle \frac { a }{ b }=\frac { 64 }{ 5 }f(0)$$
  • Question 9
    1 / -0
    If $$f(x)=\left\{\begin{matrix}(cos x)^{1/sinx} &for &x\neq 0 \\  k & for & x=0 \end{matrix}\right.$$
    Then the value of $$k$$, so that $$f$$ is continuous at $$x=0$$ is
    Solution
    $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right)  } =k$$
    $$\displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cos { x }  \right)  }^{ \cfrac { 1 }{ \sin { k }  }  } } =k$$
    $${ 1 }^{ \infty  }$$ form
    $${ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { \cos { x } -1 }{ \sin { x }  }  }  }={ e }^{\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { -\sin { x }  }{ \cos { x }  }  }  }\quad \left( \cfrac { 0 }{ 0 } form \right) $$
    $$={ e }^{ -\cfrac { 0 }{ 1 }  }=1=k$$
  • Question 10
    1 / -0
    The function $$f$$ : $$R/\{0\}\rightarrow R$$ given by $$f(x)=\displaystyle \frac{1}{x}-\frac{2}{e^{2x}-1}$$ can be made continuous at $$x=0$$ by defining $$f(0)$$ as -
    Solution
    $$f(x) = \dfrac { 1 }{ x } -\dfrac { 2 }{ { e }^{ 2x }-1 } 
    = \dfrac { { e }^{ 2x }-1-2x }{ x{ (e }^{ 2x }-1) } $$

    Using expansion of $$e^x$$

    $$f(x) = \dfrac { \left(1+2x+\dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! } +...\right)-1-2x }{ x{ (e }^{ 2x }-1) }$$
    $$f(x)=\dfrac { \dfrac { { (2x) }^{ 2 } }{ 2! } +\dfrac { { (2x) }^{ 3 } }{ 3! }  }{ x{ (e }^{ 2x }-1) } =\quad \dfrac { 2x\left(\dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! }\right) }{ x\times{ (e }^{ 2x }-1) }$$
    $$f(x)=\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac {\left(\dfrac { { (2x) } }{ 2! }
    +\dfrac { { (2x) }^{ 2 } }{ 3! } \right) }{ x } $$
    Now,
    $$\displaystyle\lim_{ x\rightarrow 0}\dfrac { 2x }{ { (e }^{ 2x }-1) }\times\dfrac { \dfrac { { (2x) } }{ 2! } +\dfrac { { (2x) }^{ 2 } }{ 3! } +... }{ x } =1\times1=1$$

    Hence, option D.
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