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Limits and Continuity Test 7

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Limits and Continuity Test 7
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  • Question 1
    1 / -0
    Evaluate: $$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$$
    Solution
    Given,

    $$\lim _{x\to \:0}\left(\dfrac{\sin \left(3\right)x^2}{\cos \left(2x^2-x\right)}\right)$$

    $$=\dfrac{\sin \left(3\right)\cdot \:0^2}{\cos \left(2\cdot \:0^2-0\right)}$$

    $$=0$$
  • Question 2
    1 / -0
    lf $$f(x)=\displaystyle \begin{cases}\dfrac{a^{2[x]+\{x\}}-1}{2[x]+\{x\}};x\neq 0 \\ \log a;x=0 \end{cases}$$ where $$[.\ ]$$ and $$\{.\ \}$$ denote integral and fractional part respectively, then
    Solution
    Given definition of $$f(x)$$ can be written as 
    $$f(x)=\displaystyle \begin{cases}\dfrac{a^{[x]+x}}{[x]+x};x\neq 0 \\ \log a;x=0 \end{cases}$$                    ($$\because \{x\}=x-[x]$$)
                               
    To check continuity of $$f(x)$$ at $$x=0$$
    $$f(0^{+})=f(0^{-})=f(0)$$
    $$f(0^{+})=\lim_{h\rightarrow 0}\dfrac{a^{h+[h]}}{\left [ h \right ]+h}$$       
    $$=\lim_{h\rightarrow 0}\dfrac{a^{h}-1}{h}=\log(a)$$

    $$f(0^-)=\lim_{h\rightarrow 0}\dfrac{a^{[-h] -h}}{[-h]-h}$$
    $$=\lim_{h\rightarrow 0}\dfrac{a^{-1-h}-1}{-1-h}$$
    $$=1-\dfrac{1}{a}$$
    $$f(0^{+})\neq f(0^{-})$$
    So, $$f(x)$$ is discontinuous at $$x=0$$
  • Question 3
    1 / -0

     lf $$\displaystyle { f }({ x })=\sqrt { \frac { { x }-\sin ^{ 2 }{ x }  }{ { x }+\cos { x }  }  } $$,then $$\displaystyle \lim _{ x\rightarrow \infty  } f(x)$$=
    Solution
    $$\displaystyle \lim _{ x\rightarrow \infty  } f(x)$$$$\displaystyle = \lim _{ x\rightarrow \infty  }\sqrt { \dfrac { { x }-\sin ^{ 2 }{ x }  }{ { x }+\cos { x }  }  }=\lim _{ x\rightarrow \infty  }\sqrt { \dfrac { 1-\dfrac{\sin^2x}{x}}{ 1+\dfrac{\cos x}{x} }}=\sqrt{\dfrac{1-0}{1-0}}=1$$
  • Question 4
    1 / -0
    The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x }  \right)  }  }{ { x }^{ 2 } }  } $$ is
    Solution
    $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( \pi \cos ^{ 2 }{ x }  \right)  }  }{ { x }^{ 2 } }  }=\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi (1-\sin ^{ 2 }{ x })  \right] }  }{ { x }^{ 2 } }  } $$
    $$\quad  =\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi-\pi\sin ^{ 2 }{ x }  \right] }  }{ { x }^{ 2 } }  }=\displaystyle \lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x }  \right] }  }{ { x }^{ 2 } }  }, [\because \sin(\pi-\theta)=\sin\theta]$$
    $$\quad \displaystyle =\pi\lim _{ x\rightarrow 0 }{ \frac { \sin { \left[ \pi\sin ^{ 2 }{ x }  \right] }  }{ \pi\sin^2x }  }\left( \frac{\sin x}{x}\right)^2 =\pi\cdot 1\cdot 1=\pi$$
  • Question 5
    1 / -0
    Let $$f(x)=\cos2x.\cot\left (\displaystyle  \frac{\pi }{4}-x \right )$$ If $$f$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ then the value of $$f(\displaystyle \frac{\pi}{4})$$ is equal to
    Solution
    $$f(x)=cos2x.cot\left ( \frac{\pi }{4}-x \right )$$ 
    Given $$f(x)$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$
    $$\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  } f(x)=f(\frac { \pi  }{ 4 } )$$
    $$\displaystyle=\lim _{ x\rightarrow \frac { \pi  }{ 4 }  } \frac { \cos { 2x }  }{ \tan { (\frac { \pi  }{ 4 } -x) }  } $$
    It is of the form $$\displaystyle \frac{0}{0}$$ , so applying L-Hospital's rule 
    $$\displaystyle=\lim _{ x\rightarrow \frac { \pi  }{ 4 }  } \frac { -2\sin { 2x }  }{ -\sec ^{ 2 }{ (\frac { \pi  }{ 4 } -x) }  } $$
    $$=2$$
  • Question 6
    1 / -0

    $$\displaystyle \lim_{x\rightarrow \infty }(\sin\sqrt{x+1}-\sin\sqrt{x})=$$
    Solution
    $$\displaystyle \lim _{ x\rightarrow \infty  }{ \left( \sin { \sqrt { x+1 }  } -\sin { \sqrt { x }  }  \right)  } $$

    $$=\displaystyle \lim _{ x\rightarrow \infty  }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x }  }{ 2 }  \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  \right)  }  }  } $$

    Now for
    $$\displaystyle \lim _{ x\rightarrow \infty  }{ { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  \right)  }  } } $$

    $$=\displaystyle \lim _{ x\rightarrow \infty  }{ \cfrac { \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  \right)  }  }{ \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  } \times \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  } $$ (sandwich theorem)

    $$=\displaystyle \lim _{ x\rightarrow \infty  }{ \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  } $$

    $$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty  }{ \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ \sqrt { x+1 } +\sqrt { x }  } \cdot \cfrac { \sqrt { x+1 } +\sqrt { x }  }{ 1 }  } $$

    $$=\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow \infty  }{ \cdot \cfrac { 1 }{ \sqrt { x+1 } +\sqrt { x }  }  } =\cfrac { 1 }{ 2 } \times 0=0$$

    Now, $$2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x }  }{ 2 }  \right)  } $$ is always finite and lies between $$\left[ -2,2 \right] $$

    $$\therefore \displaystyle \lim _{ x\rightarrow \infty  }{ 2\cos { \left( \cfrac { \sqrt { x+1 } +\sqrt { x }  }{ 2 }  \right) \sin { \left( \cfrac { \sqrt { x+1 } -\sqrt { x }  }{ 2 }  \right)  }  }  } =finite\times 0$$

    $$=0$$
  • Question 7
    1 / -0
    $$\displaystyle \lim_{x\rightarrow \infty }x\displaystyle \cos\left(\frac{\pi}{8x}\right)\sin\left(\frac{\pi}{8x}\right)=$$
    Solution
    As $$ x \rightarrow \infty   ,   \cos \left (\dfrac{\pi}{8x}\right) \rightarrow 1 $$
    Looking at the rest of the part,
    $$\underset{ x \rightarrow  \infty}{\lim}   \dfrac{\pi}{8x} \rightarrow 0 $$
    $$\underset{ x \rightarrow  \infty}{\lim}      x.\sin (\dfrac{\pi}{8x}) $$
    $$=\underset{ x \rightarrow  \infty}{\lim}       \dfrac{\sin (\dfrac{\pi}{8x})}{\dfrac{1}{x}} $$
    We multiply and divide by $$ \dfrac{\pi}{8} $$
    $$=\underset{ x \rightarrow  \infty}{\lim}  ,     \dfrac{\sin \left (\dfrac{\pi}{8x}\right)}{\dfrac{\pi}{8x}} \times \dfrac{\pi}{8x} $$
    Now using the property of a limit,
    $$\underset{ y \rightarrow 0}{\lim}  ,   \dfrac{\sin y}{y} $$ $$= 1$$
    We get, applying the limit,
    $$= 1 \times $$ $$ \dfrac{\pi}{8} $$
    $$=$$ $$ \dfrac{\pi}{8} $$
  • Question 8
    1 / -0

     $$\displaystyle \lim_{x\rightarrow\infty}\frac{\sin^{4}x-\sin^{2}x+1}{\cos^{4}x-\cos^{2}x+1}$$ is equal to
    Solution
    For all $$x$$,
    $$ {sin}^{4} x = {({sin}^{2} x)}^{2} $$
    $$=  {(1 - {cos}^{2} x)}^{2} $$
    $$=  1 - 2 {cos}^{2} x + {cos}^{4} x $$
    Thus numerator $$=  1 - 2 {cos}^{2} x + {cos}^{4} x + {cos}^{2} x = 1 - {cos}^{2} x + {cos}^{4} x $$
    Hence, for all x, numerator $$=$$ denominator.
    Thus the limit $$= 1$$ for all $$x$$
  • Question 9
    1 / -0
    $$f(x)=\left\{\begin{matrix}[x]+[-x], & \\  \lambda ,& \end{matrix}\right.\begin{matrix}x\neq 2 & \\  x=2& \end{matrix},$$ then f(x) is continuous at $$x=2$$ provided $$\lambda $$ is:
    Solution
    $$RHL=\lim_{x\rightarrow 2^{+}} [x]+[-x]$$

    $$=\lim_{h \rightarrow 0}[2+h]+[-(2+h)]$$

     $$= 2 - 3$$

    $$\Rightarrow RHL= - 1$$

    Now, $$LHL=\lim_{x\rightarrow 2^{-}} [x]+[-x]$$

    $$=\lim_{h\rightarrow 0}[2-h]+[-(2-h)] $$

    $$= 1 -2$$

    $$\Rightarrow LHL= -1$$

    Since, $$f(x)$$ is continuous at $$x=2$$

    $$LHL=RHL=f(2)$$

    $$\Rightarrow \lambda =-1$$
  • Question 10
    1 / -0

    $$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$$
    Solution
    $$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})=\lim_{x\rightarrow \infty }\frac{\sin(2^{x})}{2^x}=\frac{\mbox{Any  finite  value  between  -1  and  1}}{\infty} = 0$$
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