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Limits and Derivatives Test - 15

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Limits and Derivatives Test - 15
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  • Question 1
    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 2
    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 3
    1 / -0

    $$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$$
    Solution
    $$Put\ \theta = \dfrac{\pi }{2}-\phi $$

    $$\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}$$
  • Question 4
    1 / -0

    $$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})$$
    Solution
    $$\displaystyle \lim_{x\rightarrow \infty }2^{-x}\sin(2^{x})=\lim_{x\rightarrow \infty }\frac{\sin(2^{x})}{2^x}=\frac{\mbox{Any  finite  value  between  -1  and  1}}{\infty} = 0$$
  • Question 5
    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 6
    1 / -0
    $$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$$
    Solution
    We divide both the numerator and denominator by x.
    Since, from the property of trigonometric functions we know that $$sin x$$ and $$cos x$$ can only have their values between $$-1$$ and $$1$$
    Thus, the expression transforms to,
    $$lim _{x \rightarrow \infty }            \dfrac{2 + 7\dfrac{sin x}{x}}{4 + 3\dfrac{cos x}{x}} $$
    Now applying the limit,
    $$= \dfrac{2 +0}{4+0} $$
    $$= \dfrac{1}{2} $$
    Hence, option 'C' is correct.
  • Question 7
    1 / -0

    $$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$$
    Solution
    $$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}$$

    $$\log { k } =Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x }  } $$

    $$k={ e }^{ Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x }  }  }$$

    $$ k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \log { \sin { x }  }  }{ \cot { x }  }  }$$

    It is of the form $$\displaystyle \frac{\infty}{\infty}$$, so applying L-Hospital's rule

    $$k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \cot { x }  }{ -\csc ^{ 2 }{ x }  }  }$$

    $$k={ e }^{ 0 }=1$$
  • Question 8
    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 9
    1 / -0
    $$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$ exists and finite then $$\mathrm{a}=$$
    Solution
    $$Lt_{x \rightarrow 0}\dfrac{\sin 2x+a\sin x}{x^{3}}$$ 
    $$Lt_{ x\rightarrow 0 }\dfrac { \sin  x(2\cos { x } +a) }{ x^{ 3 } } $$
    $$=Lt_{ x\rightarrow 0 }\dfrac { (2\cos { x } +a) }{ x^{ 2 } } $$
    As $$x\rightarrow 0$$, denominator tends to 0, so the numerator also tends to 0.
    $$\Rightarrow Lt_{ x\rightarrow 0 } 2\cos { x } +a=0$$
    $$\Rightarrow a=-2$$
    Hence, option 'B' is correct.
  • Question 10
    1 / -0

    $$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$$
    Solution
    Let,    $$\text{L}= \displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}$$
    Put $$y=\pi-x\Rightarrow x\to \pi\Leftrightarrow y\to 0$$
    $$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1- 4 \tan \mathrm{(\pi-y)} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{(\pi-y)}}$$
    $$\quad \quad = \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{y}}[\because \tan (\pi-y)=-\tan y]$$
    Clearly form of the limit is $$1^{\infty}$$
    $$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}$$
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