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Limits and Continuity Test 11

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Limits and Continuity Test 11
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  • Question 1
    1 / -0
    $$f(x)=\displaystyle \frac{e^{1/x^{2}}}{e^{1/x^{2}}-1}$$ , $$x\neq 0$$, $$\mathrm{f}({0})=1$$, then $$\mathrm{f}$$ at $${x}=0$$ is:
    Solution
    Given: $$f(x)=\displaystyle \frac{e^{1/x^{2}}}{e^{1/x^{2}}-1}$$

    Divide by $$e^{1/x^2}$$ to numerator and denominator

    The function can be written in the form of,

    $$\displaystyle\lim_{ x \rightarrow 0}     \dfrac{1}{1 - {e}^{\frac{-1}{{x}^{2}}}} $$

    Applying the limit,
    $$\displaystyle f(0) = \lim_{  x \rightarrow 0}     \dfrac{1}{1 - {e}^{\frac{-1}{{x}^{2}}}}  =  1$$

    Hence, the function is continuous at $$x = 0$$

    So, the function is both left and right continuous.
  • Question 2
    1 / -0

     $$\mathrm{A}$$ function $$\mathrm{f}(\mathrm{x})$$ is defined as
    $$f(x)=\left\{ \begin{matrix} ax-b & x\leq 1 \\ 3x, & 1<x<2 \\ bx^{ 2 }-a & x\geq 2 \end{matrix} \right. $$ is continuous at
    $$x=1, 2$$ then:
    Solution
    Given $$f(x)$$ is continuous at $$x=1$$ 

    $$\displaystyle \lim _{ x\rightarrow { 1 } } f(x)=f(1)$$

    Now $$LHL=\displaystyle\lim _{ x\rightarrow { 1 }^{ - } } f(x)$$
    $$=\displaystyle\lim _{ x\rightarrow { 1 }^{ - } } ax-b$$
    $$LHL=a-b$$

    And $$f(1)=3$$

    $$\Rightarrow a-b=3$$              .....(i)

    Given $$f(x)$$ is continuous at $$x=2. $$

    $$\displaystyle \lim _{ x\rightarrow { 2 } } f(x)=f(2)$$

    Now $$LHL=\displaystyle \lim _{ x\rightarrow { 2 }^{ - } } f(x)$$
    $$=\displaystyle \lim _{ x\rightarrow { 2 }^{ - } } 3x$$
    $$LHL=6$$

    And $$f(2)=4b-a$$

    $$\Rightarrow 4b-a=6$$              .....(ii)

    Solving (i) and (ii)
    $$\Rightarrow b=3, a=6$$
  • Question 3
    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 4
    1 / -0
    The integer $$n$$ for which $$\displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)(\cos x-e^{x})}{x^{n}}$$ is finite non zero number is
    Solution
    $$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { (\cos { x } -1)\left( \cos { x } -{ e }^{ x } \right)  }{ { x }^{ n } }  } =\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { \left( \cos { x } -1 \right)  }{ { x }^{ 2 } }  \right]  } \times \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } } $$
    $$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { \cos { x } -{ e }^{ x } } }{ { x }^{ n-2 } }  } =\cfrac { 0 }{ 0 } $$ form

    $$=-\cfrac { 1 }{ 2 } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { -\sin { x } -{ e }^{ x } } }{ (n-2){ x }^{ n-3 } }  } =\cfrac { 1 }{ 2(n-2) } \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { 1 } }{ { x }^{ n-3 } }  } $$

    $$\therefore$$ for the above limit to be finite $$n-3=0$$
    $$\Rightarrow n=3$$
  • Question 5
    1 / -0
    $$\displaystyle \lim_{x\rightarrow 1}\{1-x+[x+1]+[1-x]\}$$ , where $$[x]$$ denotes greatest integer function, is
    Solution
    Substitute $$x=1+t$$
    L.H.S  $$\lim_{t\rightarrow o^{-}}  (-t+[2+t]+[-t])$$
    $$=0+1+0=1$$
    R.H.S  $$\lim_{t\rightarrow o^{+}} (-t+[2+t]+[-t])$$
    $$=0+2-1=1$$
    L.H.S$$=$$R.H.S
  • Question 6
    1 / -0
    Given that the function $$\mathrm{f}$$ is defined by $$f(x)=\left\{\begin{array}{l}2x-1,x>2\\k, x=2\\x^{2}-1,x<2\end{array}\right.$$is continuous at x = 2. Then $${k}$$ is:
    Solution
    Given $$f(2)=k$$
     
    $$RHL =\lim _{ x\rightarrow { 2 }^{ + } }2x-1$$
    $$=\lim_{h\rightarrow 0}2(2+h)-1$$
    $$=3$$
    Given, f(x) is continuous at x=2.
    $$LHL=RHL =f(2)$$
    $$\Rightarrow k=3$$
  • Question 7
    1 / -0
    lf the function $$\mathrm{f}({x})=\begin{cases}\dfrac{\sin 3x}{x} &(x\neq 0) \\ \dfrac{k}{2}&(x=0) \end{cases}$$ is continuous at $${x}=0$$, then $${k}$$ is:
    Solution
    Since $$f(x)$$ is continuous at $$x = 0$$
    LHL = RHL
    $$\underset{x \rightarrow {0}^{-}}{Lim}\  f(x) = lim x \rightarrow {0}^{+} f(x) $$
    $$\underset{x \rightarrow {0}^{-}}{Lim} \       \dfrac{sin 3x}{x}  = lim x \rightarrow {0}^{+}    \dfrac{k}{2} $$
    $$\underset{ x \rightarrow {0}^{-}}{Lim}  \     \dfrac{sin 3x}{3x} \times 3 =  lim x \rightarrow {0}^{+}    \dfrac{k}{2} $$
    $$\Rightarrow 3 =  \dfrac{k}{2} $$
    Hence, $$k = 6$$
  • Question 8
    1 / -0
    $$\displaystyle \lim_{x\rightarrow 0}\frac{1}{x}\cos ^{ -1 }{ \left( \frac { 1-x^{ 2 } }{ 1+x^{ 2 } }  \right)  } =$$
    Solution
    $$\lim_{x\rightarrow 0^{-}} \dfrac{-2}{\dfrac{2x}{(1+x^{2})}}\dfrac{sin^{-1}}{(1+x^{2})}\dfrac{(2x)}{1+x^{2}}=-2$$

    $$\lim_{x\rightarrow 0^{+}}   \dfrac{2}{(\dfrac{2x}{1+x^{2}})(1+x^{2})}  sin^{-1}(\dfrac{2x}{1+x^{2}})=2$$


    $$L.H.S\neq R.H.S$$
  • Question 9
    1 / -0
    The right-hand limit of the function $$\sec{x}$$ at $$\displaystyle x=-\frac { \pi  }{ 2 } $$ is
    Solution
    Function $$\displaystyle f\left( x \right)=\sec { x } $$ and point $$\displaystyle x=-\left( \frac { \pi  }{ 2 }  \right) $$.
    We know that right$$-$$hand limit of the function $$f(x)$$ at $$x=-\left( \frac { \pi  }{ 2 }  \right) $$ is $$\displaystyle \lim _{ x\rightarrow { \left[ -\left( \pi /2 \right)  \right]  }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { \left[ -\left( \pi /2 \right)  \right]  } }{ \sec { x }  } .$$
    Substituting $$\displaystyle x=h+\left( -\frac { \pi  }{ 2 }  \right) $$ and $$h\rightarrow 0,$$ we get
    $$\displaystyle \lim _{ h\rightarrow 0 }{ \sec { \left[ h+\left( -\frac { \pi  }{ 2 }  \right)  \right]  } = } \lim _{ h\rightarrow 0 }{ \sec { \left( \frac { \pi  }{ 2 } -h \right)  }  } =\lim _{ h\rightarrow 0 }{ \csc { h }  } =\csc { 0 } =\infty $$
    $$\displaystyle \left[ \because \sec { \left( -\theta  \right) = } \sec { \theta  }  \right] $$
  • Question 10
    1 / -0

    $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1+a^{3})+8e^{1/x}}{1+(1-b^{3})e^{1/x}}=2$$ then
    Solution
    We have,
    $$\displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + (1 - b^3) e^{1/x}} = 2$$ 
                          $$\displaystyle \left [ \frac{\infty}{\infty} form \right ]$$

    $$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) e^{-1/x} + 8}{e^{-1/x} + (1 - b^3)} = 2$$

    $$\Rightarrow \displaystyle \frac{0 + 8}{0 + (1 - b^3)} = 2 \Rightarrow 1 - b^3 = 4 \Rightarrow b^3 = - 3 \Rightarrow b = (-3)^{1/3}$$

    Again,
    $$\displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + (1 - b^3) e^{1/x}} = 2$$

    $$\Rightarrow \displaystyle \lim_{x \rightarrow 0} \frac{(1 + a^3) + 8e^{1/x}}{1 + 4e^{1/x}} = 2$$

    $$\Rightarrow 1 + a^3 = 2$$             [On comparison]

    $$\Rightarrow a = 1$$
    Therefore, $$a = 1, b = (-3)^{1/3}$$
    Hence, option 'A' is correct.
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