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Limits and Continuity Test 13

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Limits and Continuity Test 13
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  • Question 1
    1 / -0

    The value of f(0)\mathrm{f}({0}) so that the function f(x)=1+x1+x3x\displaystyle \mathrm{f}({x})=\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x} becomes continuous is equal to
    Solution
    Since, f is continuous 
    f(0)=limx01+x1+x3 x  \displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { \sqrt { 1+x } -\sqrt [ 3 ]{ 1+x }  }{ x }  } 
    It is of 00\dfrac{0}{0} form 
    By L-Hospital's rule
    f(0)=limx0(121+x 13(1+x) 2/3 ) =1213=16 \displaystyle f\left( 0 \right) =\lim _{ x\rightarrow 0 }{ \left( \frac { 1 }{ 2\sqrt { 1+x }  } -\frac { 1 }{ 3{ \left( 1+x \right)  }^{ 2/3 } }  \right)  } =\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } 
  • Question 2
    1 / -0
    If f(x)f(x) is continuous and f(92)=29f\left(\dfrac {9}{2}\right)=\dfrac {2}{9}, then limx0f(1cos3xx2)\displaystyle\lim _{ x\rightarrow 0 }f\left(\frac {1-\cos 3x}{x^2}\right) is equal to :
    Solution
    limx01cos3xx2 \displaystyle\lim _{ x\rightarrow 0 }{ \frac { 1-cos3x }{ x^{ 2 } }  }

    =limx02sin2(3x2) (3x2)2 ×94 =\lim _{ x\rightarrow 0 }{ \dfrac { 2\sin ^{ 2 }{ (\frac { 3x }{ 2 } ) }  }{ (\frac { 3x }{ 2 } )^{ 2 } }  } \times \dfrac { 9 }{ 4 }

    =92\displaystyle =\frac{9}{2}

    limx0f(1cos3xx2 ) =f(92)=29\Rightarrow \displaystyle \lim _{ x\rightarrow 0 }{ f\left( \frac { 1-cos3x }{ x^{ 2 } }  \right)  } = f\left( \dfrac 9 2\right)=\dfrac { 2}{ 9 }
     .
  • Question 3
    1 / -0
    If f(x)=xsinxx+cos2x\displaystyle f(x) = \sqrt {\frac{{x - \sin x}}{{x + {{\cos }^2}x}}} then limxf(x)\mathop {\lim }\limits_{x \to \infty } f(x)  is
    Solution
     f(x)=xsin xx+cos2x=1sin xx1+cos2xx\displaystyle f(x) =\sqrt{\frac{x-sin  x}{x+ cos^2x}} =\sqrt{\frac{1- \frac{sin  x}{x}}{1+ \frac{cos^2x}{x}}}
    limxf(x)=1\lim_{x \rightarrow \infty}f(x)=1               limxsin xx=0\displaystyle \lim_{x \rightarrow \infty} \frac{sin  x}{x}=0
                                                     limx cos2xx=0\displaystyle \lim_{x \rightarrow  \infty} \frac{cos^2 x}{x}=0
  • Question 4
    1 / -0
    limx(1x+2x+3x+.........+nx)1/x\displaystyle\lim_{x \rightarrow \infty}(1^x + 2^x + 3^x+.........+n^x)^{1/x} is
    Solution
    Consider that y=limx(1x+2x+3x+.........+nx)1/xy=\displaystyle\lim_{x\to\infty}(1^x+2^x+3^x+.........+n^x)^{1/x}
    take natural logarithm on both sides, we have
    lny=limxln[(1x+2x+3x+.........+nx)1/x]\ln y=\displaystyle\lim_{x\to\infty}\ln\left[(1^x+2^x+3^x+.........+n^x)^{1/x}\right]
    lny=limxln(1x+2x+3x+.........+nx)x\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{\ln (1^x+2^x+3^x+.........+n^x)}{x}
    RHS has \dfrac{\infty}{\infty} form. Hence, use L'hospital's rule
    lny=limx1xln1+2xln2+3xln3+.........+nxlnn1x+2x+3x+.........+nx\therefore \ln y=\displaystyle\lim_{x\to\infty}\dfrac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n }{1^x+2^x+3^x+.........+n^x}
    On approximating this form, we have
    lny=limxnxlnnnx\ln y=\displaystyle\lim_{x\to\infty}\dfrac{n^x\ln n }{n^x}
    lny=lnn\ln y=\ln n
    y=n\therefore y=n
    Hence, this is required answer.
  • Question 5
    1 / -0
    limx(2x+1)40(4x1)5(2x+3)45\displaystyle \lim_{x\to\infty}{\displaystyle \frac{(2x + 1)^{40}(4x - 1)^5}{(2x + 3)^{45}}} is equal to
    Solution

    limx(2x+1)40(4x1)5 (2x+3)45\displaystyle\lim_{x\to\infty}\frac{(2x+1)^{40}(4x-1)^5  }{(2x+3)^{45}}

    =limx(2+1x)40(41x)5(2+3x)45=\displaystyle\lim_{x\to\infty}\frac{\left(2+\Large \frac{1}{x}\right)^{40}\left(4-\Large \frac{1}{x}\right)^5}{\left(2+\Large \frac{3}{x}\right)^{45}}=24045245=\displaystyle \frac{2^{40}4^5}{2^{45}}\\

    =25=32=2^5 = 32

  • Question 6
    1 / -0
    If $$f(x) =\left\{\begin{matrix}
    \dfrac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}},x>0 & \\
    e^{x}\sin x+\pi x+\lambda \ln 4,x\leq 0 &
    \end{matrix}\right.iscontinuousatis continuous at x= 0,then, then \lambda$$ is a
    Solution
    limxhf(x)=λln4\displaystyle\lim_{x\to-h} f(x) =\lambda \ln4
    limx0+hf(x)=23x22x2x+1xx2=λln4\displaystyle\lim_{x\to 0+h} f(x)= \dfrac { 2^{ 3x }-2^{ 2x }-2^{ x }+1^{ x } }{ x^{ 2 } } =\lambda \ln4

    limh023h22h2h+1hh2=λln4\displaystyle\Rightarrow \lim_{h\to 0}\dfrac { 2^{ 3h }-2^{ 2h }-2^{ h }+1^{ h } }{ h^{ 2 } } =\lambda \ln4

    limh0(22h(2h1)1(2h1))h2=λln4\displaystyle\Rightarrow \lim_{ h\to 0}\dfrac{(2^{ 2h }(2^{ h }-1)-1(2^{ h }-1))}{h^{ 2 }}=\lambda \ln4

    limh0(2h+1)(2h1)2h2=λln4\displaystyle\Rightarrow \lim_{h\to 0}\dfrac{(2^{ h }+1){ (2^{ h }-1) }^{ 2 }}{h^{ 2} }=\lambda \ln4

    limh02(2h1)2h2=λln4\displaystyle\Rightarrow \lim_{ h\to 0} \dfrac{2{ (2^{ h }-1) }^{ 2 }}{h^{ 2 }}=\lambda \ln4

    Applying L'hospital's rule
    limh0(2h1)ln4h=λln4\displaystyle\Rightarrow \lim_{h\to 0}(2^{ h }-1)\dfrac{\ln4}{h} =\lambda \ln4
    ln2ln4=λln4\Rightarrow \ln2\ln4 = \lambda \ln4
    ln2=λ\Rightarrow \ln2=\lambda

    Hence, option 'B' is correct.
  • Question 7
    1 / -0
    The function f:(R0)f :( R-{0})  \rightarrow  R given by f(x)=1x2e2x1\displaystyle f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1} can be made continuous at x=0x = 0 by defining f(0)f(0) as
    Solution
    Givenf(x)=1x2e2x1\displaystyle f\left( x \right) =\frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 }
    f(0)=limx0{1x2e2x1 } =limx0e2x12xx(e2x1)    .......[00form] \displaystyle \Rightarrow f\left( 0 \right)=\lim _{ x\rightarrow 0 }{ \left\{ \frac { 1 }{ x } -\frac { 2 }{ { e }^{ 2x }-1 }  \right\}  } =\lim _{ x\rightarrow 0 }{ \frac { { e }^{ 2x }-1-2x }{ x\left( { e }^{ 2x }-1 \right)  }  }  \ ....... \quad \left[ \frac { 0 }{ 0 } form \right]
    \therefore  usingL'Hospital rule
    f(0)=limx02e2x2 (e2x1+2xe2x) f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad  }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) }  }
    f(0)=limx04e2x4xe2x+2e2x+2e2x =4.e04(0+e0) =1\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } }  } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right)  } =1
  • Question 8
    1 / -0
    limnn(2n+1)2(n+2)(n2+3n1)\displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n + 1)^2}{(n + 2)(n^2 + 3n - 1)}} is equal to
    Solution
    limnn(2n+1)2(n+2)(n2+3n1) \displaystyle \lim_{n\to\infty}{\displaystyle \frac{n(2n+1)^2}{(n+2)(n^2+3n-1)} }

    =limn(2+1n)2(1+2n )(1+3n1n2 )  = \displaystyle \lim_{n\to\infty}{\displaystyle \frac {\left(2+\Large \frac{1}{n} \right)^2}{\left(1+\Large \frac{2}{n}  \right)\left(1+\Large \frac{3}{n} - \Large \frac{1}{n^2}  \right)} }

    =(2+0)2(1+0)(1+0+0)=4= \displaystyle \frac{(2+0)^2}{(1+0)(1+0+0)} = 4
  • Question 9
    1 / -0
    Let f:RR\mathrm{f}:\mathrm{R}\rightarrow \mathrm{R} be defined by $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
    \mathrm{k}-2\mathrm{x}, \mathrm{i}\mathrm{f}   \mathrm{x}\leq-1\\
    2\mathrm{x}+3, \mathrm{i}\mathrm{f}   \mathrm{x}>-1
    \end{array}\right.becontinous.thenfind possiblevalueof be continous. then find possible value of \mathrm{k}$$ is
    Solution
    f(x)=k2x\mathrm{f}(\mathrm{x})=\mathrm{k}-2\mathrm{x} if x1\mathrm{x}\leq-1 

              =2x+3=2\mathrm{x}+3 if x>1\mathrm{x}>-1 

    So, L.H.L.=R.H.L.L.H.L.= R.H.L.

    k2x=2x+3k-2x=2x+3

    k=4x+3k=4x+3

    For x=1x=-1, k=1k=-1

    Hence, option C'C' is correct.
  • Question 10
    1 / -0
    If f(x)=  {x1,x12x22,x<1,g(x)= {x+1,x>0x2+1,x0f(x) = \displaystyle \left\{\begin{matrix}x - 1, & x \geq 1 \\ 2x^2 - 2, & x < 1\end{matrix}\right. , g(x) = \left\{\begin{matrix}x + 1, & x > 0 \\ -x^2 + 1, & x \leq 0\end{matrix}\right., and h(x)=xh(x) = |x|, then limx0f(g(h(x)))\displaystyle \lim_{x \rightarrow 0} f(g (h (x))) is
    Solution
    limx0+f(g(h(x)))=f(g(0+))=f(1+)=0\displaystyle \lim_{x \rightarrow 0^+} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0
    limx0f(g(h(x)))=f(g(0+))=f(1+)=0\displaystyle \lim_{x \rightarrow 0^-} f(g(h(x))) = f(g(0^+)) = f(1^+) = 0
    Therefore, limx0f(g(h(x)))=0\displaystyle \lim_{x \rightarrow 0} f(g(h(x))) = 0
    Hence, option 'A' is correct.
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