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Limits and Continuity Test 17

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Limits and Continuity Test 17
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  • Question 1
    1 / -0
    The value of the constant $$\alpha$$ and $$\beta$$ such that $$\displaystyle \lim_{x\rightarrow \infty}\left(\displaystyle\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$$ are respectively.
    Solution
    Simplifying the above expression gives us

    $$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}+1-\alpha x(x+1)-\beta(x+1)}{x+1})$$

    $$=\displaystyle \lim_{x\rightarrow \infty}(\dfrac{x^{2}(1-\alpha)-x(\alpha +\beta)+1-\beta)}{x+1})$$
    $$=0$$ implies that $$Numerator<Denominator$$.
    Hence coefficient of $$x^{2}$$ will be 0. Therefore $$\alpha=1$$.
    Hence
    $$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-x(1 +\beta)+1-\beta)}{x+1})=0$$

    Applying L'Hopital's Rule we get
    $$\displaystyle \lim_{x\rightarrow \infty}(\dfrac{-(1 +\beta)}{1})=0$$
    Or
    $$\beta+1=0$$
    Or
    $$\beta=-1$$.
    Hence
    $$(\alpha,\beta)=1,-1$$
  • Question 2
    1 / -0
    The function represented by  the following graph is.

    Solution
    Since there is no discontinuity in the graph so clearly function is continuous,
    but is not differentiable at $$x=1$$, because graph of the function has a sharp point or kink.

    Note: The point on the graph of any function where it has a sharp corner or kink, function is not differentiable .
  • Question 3
    1 / -0
    If [x] denotes the greatest integer not exceeding $$x$$ and if the function $$f$$ defined by $$f(x)= \begin{cases}\dfrac{a+2\cos\,x}{x^2}&(x < 0) \\ b\,\tan \dfrac{\pi}{[x+4]}&(x \ge 0) \end{cases}$$ is continuous at $$x=0$$, then the order pair (a, b) =
    Solution
    Since the function is continuous at $$x = 0, f(0^-) = f(0^+)$$

    $$\Rightarrow \displaystyle \lim_ {x \rightarrow 0^-} \cfrac{a + 2\cos{x}}{x^2} = \displaystyle \lim_{x \rightarrow 0^+} b \ \tan \ \cfrac{\pi}{[x + 4]}$$ 
    For a finite limit, $$a$$ has to be $$-2$$ since the numerator has to be 

    $$0$$ for the limit to be of the form $$\dfrac{0}{0}$$
    Using L'Hospital rule, we get $$\displaystyle \lim_{x \rightarrow 0^-} \cfrac{-2 \ \sin{x}}{2x} = -1$$

    $$\Rightarrow b \ \tan \ \cfrac{\pi}{4} = -1$$

    $$\Rightarrow b = -1$$
  • Question 4
    1 / -0
    $$\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}}$$ is ____
    Solution
    $$\displaystyle \lim_{x\rightarrow 0}\dfrac {1 - \cos x}{x^{2}}=\lim_{x\to 0}\dfrac{2\sin^2\frac{x}{2}}{x^2}$$

    $$=\displaystyle \lim_{x\rightarrow 0}\dfrac {2\sin^2\frac{x}{2}}{4\left(\frac{x}{2}\right)^{2}}=\dfrac{1}{2}\lim_{x\to 0}\left( \dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\dfrac{1}{2}\cdot 1=\dfrac{1}{2}$$ (using basic limit formula )
  • Question 5
    1 / -0
    The limit of $$\left[\frac{1}{x^2}+\frac{(2013)^x}{e^x-1}-\frac{1}{e^x-1}\right]$$ as $$x\rightarrow 0$$
    Solution
    $$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right)  }^{ x }-1 }{ { e }^{ x }-1 }  \right]  } $$
    $$=\displaystyle \lim _{ x\rightarrow 0 }{ \left[ \cfrac { 1 }{ { x }^{ 2 } } +\cfrac { { \left( 2013 \right)  }^{ x }-1 }{ x } \times \cfrac { x }{ { e }^{ x }-1 }  \right]  } $$

    $$=\left( \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1 }{ { x }^{ 2 } }  }  \right) +\left( \log { 2013 }  \right) \times 1\quad \quad \left[ \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { a }^{ x }-1 }{ x }  } =\log _{ e }{ a } ;\quad \displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { { e }^{ x }-1 }{ x }  } =1 \right] $$

    $$=\infty +\log { 2013 } =\infty $$

    as $$x\rightarrow 0\Rightarrow $$Limit $$\rightarrow \infty $$
  • Question 6
    1 / -0
    If the function $$f(x)$$ satisfies $$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$$, then $$\displaystyle \lim_{x\rightarrow 1}f(x)=$$
    Solution
    It is given that, $$\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-2}{x^2-1}=\pi$$ exists and equal to $$1$$.

    Clearly the denominator is zero at $$x=1$$, so for above limit to exist 
    numerator should also be zero, as x approaches $$1$$.
    Therefore, $$\displaystyle \lim_{x\to 1}f(x)=2$$

    Note: If $$\displaystyle \lim_{x\to 1}f(x) \neq 1$$, given limit will not exist 
  • Question 7
    1 / -0
    $$\displaystyle \lim_{x\rightarrow 0}\frac{log_e(1+x)}{3^x-1}=$$ ____.
    Solution
    Given that: $$\displaystyle \lim_{x \rightarrow 0} \cfrac{\log_e(1 + x)}{3^x - 1}$$
    Using L'Hospital rule, we differentiate the numerator and denominator to find the limit.

    We thus get $$\displaystyle \lim_{x \rightarrow 0} \cfrac{1}{(1 + x)(3^x)(\log_e3)} $$

    $$= \cfrac{1}{1 \times 1 \times \log_e3}$$$$= \cfrac{1}{\log_e3}$$$$= \log_3e$$
  • Question 8
    1 / -0
    The limit of $$\displaystyle \sum_{n=1}^{1000}(-1)^nx^n$$ as $$x\rightarrow \infty$$
    Solution
    $$\sum _{ n=1 }^{ 1000 }{ { \left( -1 \right)  }^{ n }{ x }^{ n } } =-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }+....$$
    It is in GP
    $$S=\cfrac { -x\left( { (-x) }^{ 1000 }-1 \right)  }{ (-x)-1 } $$

    $$\displaystyle \lim _{ x\rightarrow \infty  }{ \sum _{ n=1 }^{ 1000 }{ { \left( -1 \right)  }^{ n }{ x }^{ n } }  } =\displaystyle \lim _{ x\rightarrow \infty  }{ S } =\displaystyle \lim _{ x\rightarrow \infty  }{ \cfrac { x\left( { x }^{ 1000 }-1 \right)  }{ x+1 }  } $$

    $$=\displaystyle \lim _{ x\rightarrow \infty  }{ \cfrac { \left( { x }^{ 1000 }-1 \right)  }{ 1+\cfrac { 1 }{ x }  }  } =\cfrac { { \infty  }^{ 1000 }-1 }{ 1+0 } =\infty $$ 

     $$as \left( x\rightarrow \infty ;\cfrac { 1 }{ x } \rightarrow \infty \right) $$

  • Question 9
    1 / -0
    The limit of $$\left\{\frac{1}{x}\sqrt{1-x}-\sqrt{1+\frac{1}{x^2}}\right\}$$ as $$x\rightarrow 0$$
    Solution
    Let $$p(x)=\cfrac { 1 }{ x } \sqrt { 1-x } ,q(x)=\sqrt { 1+\cfrac { 1 }{ { x }^{ 2 } }  } $$
    $$\lim _{ x\rightarrow { 0 }^{ + } }{ p\left( x \right)  } =\lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x } \sqrt { 1-x }  } =\infty \times \sqrt { 1-10 } $$
    $$=\infty $$
    $$\lim _{ x\rightarrow { 0 }^{ - } }{ p\left( x \right)  } =-\infty \times \sqrt { 1-0 } =-\infty $$
    as $$x\rightarrow { 0 }^{ + }\quad \cfrac { 1 }{ x } \rightarrow +\infty $$
    $$x\rightarrow { 0 }^{ - }\quad \cfrac { 1 }{ x } \rightarrow -\infty $$
    $$\Rightarrow p(x)$$ is discontinuous as $$x\rightarrow 0$$
    Let $$f\left( x \right) =p\left( x \right) -q\left( x \right) $$
    Since $$p\left( x \right) $$ is discontinuous $$f\left( x \right) $$ is also discontinuous by Algebra of continuous functions.
  • Question 10
    1 / -0
    $$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x}$$ is equal to
    Solution
    $$\displaystyle \lim_{x\rightarrow \pi/4} \dfrac {\tan x - 1}{\cos 2x} = \displaystyle \lim_{h\rightarrow 0}\dfrac {\tan \left (\dfrac {\pi}{4} + h\right ) - 1}{\cos 2\left (\dfrac {\pi}{4} + h\right )}$$
    $$\left [\because x = \dfrac {\pi}{4} + h\right ]$$

    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {\left (\dfrac {1 + \tan h}{1 - \tan h}\right ) - 1}{\cos \left (\dfrac {\pi}{2} + 2h\right )}$$

    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {1 + \tan h - 1 + \tan h}{-\sin 2h (1 - \tan h)}$$
    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-2\tan h}{2\sin h \cos h (1 - \tan h)}$$
    $$= \displaystyle \lim_{h\rightarrow 0} \dfrac {-1}{\cos^{2} h (1 - \tan h)} = -1$$
    Hence, option D is correct.
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