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Limits and Continuity Test 22

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Limits and Continuity Test 22
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  • Question 1
    1 / -0
    limxπ2cotxcosx(π2x)3=\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} =
    Solution

  • Question 2
    1 / -0
    limx01xsin1(2x1+x2)\mathop {\lim }\limits_{x \to 0} \,\dfrac{1}{x}\,{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) is equal to
    Solution
    Solution -
    limx01xsin1(2x1+x2) \displaystyle \lim_{x\rightarrow 0} \frac{1}{x}sin^{-1}(\frac{2x}{1+x^{2}})
    we know that
    sin1x+sin1y=sin1(x1y2+y1x2) \displaystyle sin^{-1}x+sin^{-1}y = sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})
    sin1(2x1+x2)=sin1(11+x2)+sin1(11+x2) \displaystyle sin^{-1}(\frac{2x}{1+x^{2}})= sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})+sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})
    =2sin1(11+x2)π \displaystyle = 2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi
    limx02sin1(11+x2)πx\displaystyle \lim_{x\rightarrow 0} \frac{2sin^{-1}(\frac{1}{\sqrt{1+x^{2}}})-\pi }{x}
    from triangle.
    limx02tan1(1x)πx \displaystyle \lim_{x\rightarrow 0} \frac{2tan^{-1}(\frac{1}{x})-\pi }{x}
    Apply L Hospital rule-
    limx02(11+(1x)2) \displaystyle \lim_{x\rightarrow 0} 2(\frac{1}{1+(\frac{1}{x})^{2}})
    limx02x21+x2 \displaystyle \lim_{x\rightarrow 0} \frac{2x^{2}}{1+x^{2}}
    =0. = 0.
    B is correct.

  • Question 3
    1 / -0
    solve the limit 
    limx32x3\mathop {\lim }\limits_{x \to 3} \dfrac{2}{{x - 3}} 
    Solution
    limx32x3\displaystyle \lim_{x\rightarrow 3} \dfrac{2}{x-3}
    put h=x3h = x - 3
    limh02h\displaystyle \lim_{h\rightarrow 0} \dfrac{2}{h}
     this is 20\dfrac{2}{0} that is not defined. 
    so limit does not exist.
  • Question 4
    1 / -0
    Solve

    limx0sin5xtan3x\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 5x}}{{\tan 3x}}
    Solution
    LetL=limx0sin5xtan3xL=\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}

    =limx0sin5x5x×3xtanx×53= \displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}\times \dfrac {3x}{\tan x}\times \dfrac {5}{3}

    we know that limx0sin5x5x=1\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{5x}=1 and limx03xtan3x=1\displaystyle \lim_{x\rightarrow 0}\dfrac {3x}{\tan 3x}=1


    L=1×1×53L=1\times 1\times \dfrac {5}{3}

    limx0sin5xtan3x=53\displaystyle \lim_{x\rightarrow 0}\dfrac {\sin 5x}{\tan 3x}= \dfrac {5}{3}


  • Question 5
    1 / -0
    limx01xsin1(2x1+x2)\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) is equal to
    Solution
    limx01xsin1(2x1+x2)\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)

    can be written as 
    limx0sin1(2x1+x2)x\mathop {\lim }\limits_{x \to 0} \dfrac{\sin^{-1}(\dfrac{2x}{1+x^2})}{x}

    it is of the form 00\dfrac00

    So, we will apply L Hospital rule until we will get determinate form

    Differentiating numerator and denominator we will get,
    =limx011(2x1+x2)2.2(1+x2)4x2(1+x2)21=limx011(2x1+x2)2.2(1+x2)4x2(1+x2)2=2=\lim_{x\to 0}\dfrac{\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}}{1}=\lim_{x\to0}\dfrac{1}{\sqrt{1-(\dfrac{2x}{1+x^2})^2}}.\dfrac{2(1+x^2)-4x^2}{(1+x^2)^2}=2
  • Question 6
    1 / -0
    limx1(2x3)(x 1)2x2+x3=\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{2{x^2} + x - 3}} =
    Solution
    Now,
    limx1(2x3)(x 1)2x2+x3 \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{2{x^2} + x - 3}} 

    =limx1(2x3)(x 1)(2x2+3x2x3) =\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{(2{x^2} +3 x-2x - 3)}} 

    =limx1(2x3)(x 1)(x1)(2x+3) =\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{(x-1)(2x + 3)}} 

    =limx1(2x3)(x 1)(x1)(x+1)(2x+3) =\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{(\sqrt x-\sqrt 1)(\sqrt x+\sqrt 1)(2x + 3)}} 

    =limx1(2x3)(x+1)(2x+3) =\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)}}{{(\sqrt x+\sqrt 1)(2x + 3)}} 

    =12×5=\dfrac{-1}{2\times5}

    =110=-\dfrac{1}{10}.
  • Question 7
    1 / -0
    If   zr=cosrαn2+isinrαn2{z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}, where r=1,2,3,....n r= 1, 2, 3, ....n, then limn(z1.z2.....zn)\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right) is equal to 
    Solution
    Solution - 
    zr=cosrαn2+isinrαn2 z_{r} = \dfrac{cosr\alpha }{n^{2}}+i\dfrac{sinr\alpha }{n^{2}}

    Comparing with cosθ+isinθ cos \theta +isin \theta

    θ=rαn2\theta =\dfrac {r\alpha}{n^2}

    zr=eirαn2\therefore z_r= e^{\dfrac{ir\alpha }{n^{2}}}

    limn(z1,z2,z3....zn)=eiαn2e2iαn2e3iαn2...eniαn2 \lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = e^{\dfrac{i\alpha }{n^{2}}}e^{\dfrac{2i\alpha }{n^{2}}} e^{\dfrac{3i\alpha }{n^{2}}}...e^{\dfrac{ni\alpha }{n^{2}}}

    =limn eiα(1n2+2n2+...) = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{1}{n^{2}}+\dfrac{2}{n^{2}}+...)}

    =limneiα1n2(1+2+3) = \lim_{n\rightarrow \infty } e^{i\alpha \dfrac{1}{n^2}(1+2+3)}

    =limn eiα(n2+n2n2)Sn=n2[2a+(n1)d] = \lim_{n\rightarrow \infty } e^{i\alpha (\dfrac{n^{2}+n}{2n^{^{2}}})} \quad \dots S_{n}=\dfrac{n}{2}\left [ 2a+(n-1)d \right ]

    at n n\rightarrow \infty

    =eiα×12 = e^{i\alpha }\times \dfrac{1}{2}

    limn(z1,z2,z3....zn)=eiα\lim_{n\rightarrow \infty } (z_{1},z_{2},z_{3}....z_n) = \sqrt{e^{i\alpha }}



  • Question 8
    1 / -0
    Solve:
    01dxx+1+xdx=\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}dx=
    Solution
    01dxx+1+x\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}}

    =01dxx+1+x×x+1xx+1x=\displaystyle\int_{0}^{1}{\dfrac{dx}{\sqrt{x+1}+\sqrt{x}}\times\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}}

    =01(x+1+x)dxx+1x=\displaystyle\int_{0}^{1}{\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}}

    =01(x+1+x)dx=\displaystyle\int_{0}^{1}{\left(\sqrt{x+1}+\sqrt{x}\right)dx}

    =[(x+1)12+112+1+(x)12+112+1]01=\left[\dfrac{{\left(x+1\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}+\dfrac{{\left(x\right)}^{\frac{1}{2}+1}}{\dfrac{1}{2}+1}\right]_{0}^{1}

    =[2(x+1)323+2(x)323]01=\left[2\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+2\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}

    =2[(x+1)323+(x)323]01=2\left[\dfrac{{\left(x+1\right)}^{\frac{3}{2}}}{3}+\dfrac{{\left(x\right)}^{\frac{3}{2}}}{3}\right]_{0}^{1}

    =23[(2321)(10)]=\dfrac{2}{3}\left[\left({2}^{\frac{3}{2}}-1\right)-\left(1-0\right)\right]

    =23[222]=\dfrac{2}{3}\left[2\sqrt{2}-2\right]

    =43(21)=\dfrac{4}{3}\left(\sqrt{2}-1\right)
  • Question 9
    1 / -0
    If f(x)f(x) is the integral of 2sinxsin2xx3, x0\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0. Find limx0f(x) \lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right)  } , where f(x)=df(x)dxf^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}
    Solution
    f(x)=2sinxsin2xx3f(x)=\dfrac{2\sin x-\sin 2x}{x^3}    x0x\neq 0

    f(x)=df(x)dx=2sinxsin2xx3f'(x)=\dfrac{df(x)}{dx}=\dfrac{2\sin x-\sin 2x}{x^3}

    =2sin xx(1cosxx2)=\dfrac{2\sin  x}{x}\left(\dfrac{1-\cos x}{x^2}\right)

    limx0f(x)=limx02(sinxx)(2sin2(x/2)x2)\Rightarrow \displaystyle \lim_{x\rightarrow 0}{f'(x)}=\displaystyle \lim_{x\rightarrow 0}{2\left(\dfrac{\sin x}{x}\right)}\left( \dfrac{2\sin^2(x/2)}{x^2}\right)

    41limx0(sin2(x/2)4×(x/2)2)\Rightarrow \displaystyle 4\cdot 1 \lim_{x\rightarrow 0}{\left(\dfrac{\sin^2(x/2)}{4\times (x/2)^2}\right)}

    =1=1
  • Question 10
    1 / -0
    Find the value of limit limxπ 6 2sin2x+sinx1 2sin2x3sinx+1 = \displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 }  }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 }  } = } .
    Solution
    limxπ 6 2sin2x+sinx12sin2x3sinx+1 =limxπ 6 2sin2x+2sinxsinx12sin2x2sinxsinx+1 =limxπ 6 2sinx(sinx+1)1(sinx+1) 2sinx(sinx1)1(sinx1)  =limxπ 6 (2sinx1)(sinx+1) (2sinx1)(sinx1)  =limxπ 6 sinx+1sinx1 =12+1121=3\lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -3\sin { x } +1 }  } \\ =\lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin ^{ 2 }{ x } +2\sin { x } -\sin { x } -1 }{ 2\sin ^{ 2 }{ x } -2\sin { x } -\sin { x } +1 }  } \\ =\lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin { x } \left( \sin { x } +1 \right) -1\left( \sin { x } +1 \right)  }{ 2\sin { x } \left( \sin { x } -1 \right) -1\left( \sin { x } -1 \right)  }  } \\ =\lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { \left( 2\sin { x } -1 \right) \left( \sin { x } +1 \right)  }{ \left( 2\sin { x } -1 \right) \left( \sin { x } -1 \right)  }  } \\ =\lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { \sin { x } +1 }{ \sin { x } -1 }  } \\ =\frac { \frac { 1 }{ 2 } +1 }{ \frac { 1 }{ 2 } -1 } \\ =-3
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