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Limits and Continuity Test 21

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Limits and Continuity Test 21
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  • Question 1
    1 / -0
    $$\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^{4x}} - 1}}{x}$$
    Solution

    We have,

    $$\underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{{{e}^{4x}}-1}{x} \right)$$

     

    This is $$\dfrac{0}{0}$$ form.

     

    So, by using L-Hospital rule

    $$ \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{{{e}^{4x}}\times 4-0}{1} \right) $$

    $$ \underset{x\to 0}{\mathop{\lim }}\,\left( 4{{e}^{4x}} \right) $$

    $$ =4\times {{e}^{0}} $$

    $$ =4\times 1 $$

    $$ =4 $$

     

    Hence, this is the answer.

  • Question 2
    1 / -0
    $$\displaystyle \lim_{n \rightarrow \infty}\dfrac {1^{2}+2^{2}+3^{2}+....+n^{2}}{n^{3}}$$ is equal to
    Solution
    Solution
    $$ \displaystyle \lim_{x\rightarrow \infty } \dfrac{1^2 + 2^2 +--- n^2}{n^3}$$
    we know that $$ 1^2 + 2^2 + --- n^2 = \dfrac{n (n+1)(2n+1)}{6}$$
    $$ \displaystyle =\lim_{x\rightarrow \infty }\frac{n(n+1)(2n+1)}{6n^3}$$
    $$ \displaystyle= \lim_{x\rightarrow \infty }\frac{(n+1)(2n+1)}{6n^2} = \lim_{x\rightarrow \infty } \frac{2n^2 +3n +1}{6n^2}$$
    Apply L-Hospital rule twice.
    $$ \displaystyle = \lim_{x\rightarrow \infty } \dfrac{4n+3}{12n}$$
    $$ = \dfrac{4}{12} = \dfrac{1}{3}$$
    C is correct.

  • Question 3
    1 / -0
    $${ x }_{ n }={ \left( 1-\cfrac { 1 }{ 3 }  \right)  }^{ 2 }{ \left( 1-\cfrac { 1 }{ 6 }  \right)  }^{ 2 }{ \left( 1-\cfrac { 1 }{ 10 }  \right)  }^{ 2 }...{ \left( 1-\cfrac { 1 }{ \cfrac { n(n+1) }{ 2 }  }  \right)  }^{ 2 },n\ge 2$$. Then the value of $$\displaystyle\lim _{ n\rightarrow \infty  }{ { x }_{ n } } $$
    Solution

  • Question 4
    1 / -0
    $$ \lim _{ x\rightarrow 1 }{ \dfrac { \sqrt { 1-\cos { 2\left( x-1 \right)  }  }  }{ x-1 }  }$$
    Solution

  • Question 5
    1 / -0
    The value of k which makes $$f(x)=\left\{\begin{matrix} \sin\dfrac{1}{x}, x\neq 0\\ k, x=0\end{matrix}\right.$$ continuous at $$x=0$$ is?
    Solution
    Then k=?
    $$\begin{matrix} \Rightarrow f\left( 0 \right) =k \\ \Rightarrow { f\left( { { 0^{ + } } } \right)  }_{ \lim  \, \, x\to { 0^{ + } } }={ f\left( { { 0^{ +h } } } \right)  }_{ \lim  \, \, h\to 0 }={ \sin   }_{ \lim  \, \, h\to 0 }\dfrac { 1 }{ { (o+h) } } =\sin  0 \\ \Rightarrow { f\left( { { 0^{ - } } } \right)  }_{ \lim  \, \, x\to { 0^{ - } } }={ f\left( { { 0^{ -h } } } \right)  }_{ \lim  \, \, h\to 0 }={ \sin   }_{ \lim  \, \, h\to 0 }\dfrac { 1 }{ { (o-h) } } =\sin  0 \\ \Rightarrow f\left( 0 \right) =f\left( { { 0^{ + } } } \right) =f\left( { { 0^{ - } } } \right) =f\left( 0 \right) =k=0 \\  \end{matrix}$$
    the value of $$k$$ is zero 
  • Question 6
    1 / -0
    If $$f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$$, then $$\underset{a \rightarrow 0}{\lim} \dfrac{f(1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is 
    Solution
    $$\lim_{\alpha \rightarrow 0}\dfrac{f(1-\alpha )-f(1)}{\alpha ^3+3\alpha }$$

    Apply L-Hospital's rule

    $$=\lim_{\alpha \rightarrow 0}\dfrac{f'(1-\alpha )-0}{3\alpha ^2+3 }$$

    $$=\dfrac{f'(1 )}{3 }$$

    Considering, f(x)

    $$f(x )=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$$

    $$f'(x )=30x^{9}-56x^7+30x^5-63x^2+6x$$

    Calculating required limit value:
    $$=\dfrac{f'(1 )}{3 }$$

    $$=\dfrac{30(1)^{9}-56(1)^7+30(1)^5-63(1)^2+6(1)}{3 }$$

    $$=\dfrac{-53}{3}$$
  • Question 7
    1 / -0
    Let $$f(x) = \underset{0}{\overset{x}{\int}} |2t - 3| dt$$, then $$f$$ is 
  • Question 8
    1 / -0
    If $$f(x) = 2x^9 - 5x^8 + 7x^6 - 15x^4 + 5x + 7$$, then $$\underset{x \rightarrow 0}{\lim} \dfrac{f (1 - \alpha) - f(1)}{\alpha^3 + 3 \alpha}$$ is 
  • Question 9
    1 / -0
    The value of $$\lim_{x\rightarrow o}\dfrac{\sqrt{\dfrac{1}{2}(1-cos 2 x)}}{x}$$
    Solution
    $$\lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \cos{2x} \right)}}{x}}$$
    $$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - \left( 1 - 2 \sin^{2}{x} \right) \right)}}{x}}$$
    $$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 1 - 1 + 2 \sin^{2}{x} \right)}}{x}}$$
    $$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\frac{1}{2} \left( 2 \sin^{2}{x} \right)}}{x}}$$
    $$= \lim_{x \rightarrow 0}{\cfrac{\sqrt{\sin^{2}{x}}}{x}}$$
    $$= \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}}$$
    $$= 1 \; \left( \because \lim_{x \rightarrow 0}{\cfrac{\sin{x}}{x}} = 1 \right)$$

  • Question 10
    1 / -0
    The value of $$\underset { x\longrightarrow \infty  }{ Lim } \dfrac{d}{dx}\overset { \sqrt { 3 }  }{ \underset { -\sqrt { 3 }  }{ \int }  } \dfrac{r^3}{(r+1)(r-1)}$$dr,is
    Solution
    Given,

    $$\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{(r+1)(r-1)}dr$$

    using long division process, we get,

    $$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}\dfrac{r^3}{r^2-1}dr$$

    $$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\int_{-\sqrt{3}}^{\sqrt{3}}r+\dfrac{r}{r^2-1}dr$$

    $$=\lim_{x\rightarrow \infty }\dfrac{d}{dx}\left [ \dfrac{r^2}{2}+\dfrac{1}{2}\ln \left|r^2-1\right| \right ]_{-\sqrt{3}}^{\sqrt{3}}$$

    $$=\lim_{x\rightarrow \infty }\dfrac{d}{dx} (0)$$

    $$=0$$
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