Tag: differencial calculus - limits and continuity

Questions Related to differencial calculus - limits and continuity

Let $f\left( x \right) =\dfrac { \log { \left( 1+x+{ x }^{ 2 } \right)  } +\log { \left( 1-x+{ x }^{ 2 } \right)  }  }{ \sec { x } -\cos { x }  } ,x\neq 0$ The value of $f\left (0\right)$ so that $f$ is continuous at $x=0$ is 

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $1$

  2. $0$

  3. $2$

  4. None of these

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Correct Option: A

Given $f(x)=\dfrac{\left[ \left{ \left| x \right|  \right}  \right] { e }^{ { x }^{ 2 } }\left{ \left[ \left| x+\left{ x \right}  \right|  \right]  \right} }{\left( { e }^{ 1/{ x }^{ 2 } }-1 \right) sgn\left( \sin { x }  \right) }$ for $x\neq 0$
$=0, for\  x=0$
Where $\left{ x \right} $ is the fractional part function; $[x]$ is the step up function and $sgn{(x)}$ is the signum function of $x$ then, $f(x)$

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. is continuous at $x=0$

  2. is discontinuous at $x=1$

  3. has a removable discontinuity at $x=0$

  4. has an irremovable discontinuity at $x=0$

Show answer
Correct Option: A
Explanation:
$\dfrac { \left[ \left\{ \left| x \right|  \right\}  \right] { e }^{ { x }^{ 2 } }\left\{ \left[ \left| x+\left\{  \right\}  \right|  \right]  \right\}  }{ \left( { e }^{ \dfrac { 1 }{ { x }^{ 2 } }  }-1 \right) sgn\left( \sin { x }  \right)  } $
$f(o)=\dfrac{continous}{continuous}=continuous$
$f(o)=continuous$
$f(x)$ is continuous at $x=0$

Given that $\displaystyle \prod _{n=1}^n cos \dfrac{x}{2^n}= \dfrac{\sin  x}{2^n  \sin \left ( \dfrac{x}{2^n} \right )}$ and $\displaystyle f(x) = \left{\begin{matrix}\lim _{n \rightarrow \infty}\sum _{n = 1}^n \dfrac{1}{2^n} \tan \left (\dfrac{x}{2^n} \right ), & x \in (0, \pi) - \left {\dfrac{\pi}{2} \right }\ \dfrac{2}{\pi} & x = \dfrac{\pi}{2}\end{matrix}\right.$
Then which one of the following is true?

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $f(x)$ has non-removable discontinuity of finite type at $\displaystyle x = \dfrac{\pi}{2}$.

  2. $f(x)$ has removable discontinuity at $\displaystyle x = \dfrac{\pi}{2}$

  3. $f(x)$ is continuous at $\displaystyle x = \dfrac{\pi}{2}$.

  4. $f(x)$ has non-removable discontinuity of infinite type at $\displaystyle x = \dfrac{\pi}{2}$

Show answer
Correct Option: C
Explanation:
We know that $\tan { x } =\cfrac { 2\tan { \left( \cfrac { x }{ 2 }  \right)  }  }{ 1-\tan ^{ 2 }{ \left( \cfrac { x }{ 2 }  \right)  }  } $
$\Rightarrow \cot { x } =\cfrac { 1-\tan ^{ 2 }{ \left( \cfrac { x }{ 2 }  \right)  }  }{ 2\tan { \left( \cfrac { x }{ 2 }  \right)  }  } =\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cfrac { 1 }{ 2 } \tan { \left( \cfrac { x }{ 2 }  \right)  } $
$\Rightarrow \cfrac { 1 }{ 2 } \tan { \left( \cfrac { x }{ 2 }  \right)  } =\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cot { x } \rightarrow 1$
Replacing x by $\cfrac { x }{ 2 } $ and multiplying by $\cfrac { 1 }{ 2 } $
$\cfrac { 1 }{ 4 } \tan { \left( \cfrac { x }{ 4 }  \right)  } =\cfrac { 1 }{ 4 } \cot { \left( \cfrac { x }{ 4 }  \right)  } -\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } $
Repeating the steps
$\cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } =\cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cfrac { 1 }{ { 2 }^{ n-1 } } \cot { \left( \cfrac { x }{ { 2 }^{ n-1 } }  \right)  } $
$\therefore \sum _{ n=1 }^{ n }{ \left[ \cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  \right]  } =\left[ \cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  } -\cot { x }  \right] +\left[ \cfrac { 1 }{ 4 } \cot { \left( \cfrac { x }{ 4 }  \right)  } -\cfrac { 1 }{ 2 } \cot { \left( \cfrac { x }{ 2 }  \right)  }  \right] +....+\left[ \cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cfrac { 1 }{ { 2 }^{ n-1 } } \cot { \left( \cfrac { x }{ { 2 }^{ n-1 } }  \right)  }  \right] $
$=\cfrac { 1 }{ { 2 }^{ n } } \cot { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  } -\cot { x } =\cfrac { \left( \cfrac { 1 }{ { 2 }^{ n } }  \right)  }{ \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  } -\cot { x } $
$f\left( x \right) =\lim _{ n\rightarrow \infty  }{ \sum _{ n=1 }^{ n }{ \left[ \cfrac { 1 }{ { 2 }^{ n } } \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  \right]  }  } =\lim _{ n\rightarrow \infty  }{ \left( \cfrac { \left( \cfrac { x }{ { 2 }^{ n } }  \right) \times \cfrac { 1 }{ x }  }{ \tan { \left( \cfrac { x }{ { 2 }^{ n } }  \right)  }  } -\cot { x }  \right)  } $
as $n\rightarrow \infty ,\cfrac { 1 }{ { 2 }^{ n } } \rightarrow 0$
$\cfrac { x }{ { 2 }^{ n } } \rightarrow 0$
$f\left( x \right) =\cfrac { 1 }{ x } \times 1-\cot { x } $
$x=\cfrac { \pi  }{ 2 } $
$f\left( \cfrac { \pi  }{ 2 }  \right) =\cfrac { 2 }{ \pi  } $ (given )$\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ f\left( x \right)  } =\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ \left( \cfrac { 1 }{ x } -\cot { x }  \right)  } $
$=\cfrac { 1 }{ \cfrac { \pi  }{ 2 }  } -0=\cfrac { 2 }{ \pi  } $
$\therefore f\left( \cfrac { \pi  }{ 2 }  \right) =\lim _{ x\rightarrow \cfrac { \pi  }{ 2 }  }{ f\left( x \right)  } $
$\Rightarrow f\left( x \right) $ is continuous at $x=\cfrac { \pi  }{ 2 }$

The value of f(0) so that the function
$f(x)=\displaystyle \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}$
becomes continuous, is equal to

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $\dfrac{1}{6}$

  2. $\dfrac{1}{4}$

  3. 2

  4. $\dfrac{1}{3}$

Show answer
Correct Option: A
Explanation:

The function $f(x)$ is continuous except possibly at $x=0$. For $f$ to be continuous at $x=0$, we must have
$\displaystyle f(0)=\lim _{x\rightarrow 0}f(x)=\lim _{x\rightarrow 0}\frac{(1+x)^{1/2}-(1+x)^{1/3}}{x}$

$=\displaystyle \lim _{x\rightarrow 0}\dfrac{\left [ 1+\dfrac{1}{2}x+\dfrac{(1/2)(-1/2)}{2}x^{2}+... \right ]-\left [ 1+\dfrac{1}{3}x+\dfrac{(1/3)(-2/3)}{2}x^{2}+... \right ]}{x}$
Therefore,
$\displaystyle f(0)=\lim _{x\rightarrow 0}x\dfrac{\left [ \dfrac{1}{2}+\dfrac{(1/2)(-1/2)}{2}x+... \right ]-\left [\dfrac{1}{3}+\dfrac{(1/3)(-2/3)}{2}x+... \right ]}{x}$


$=\displaystyle \lim _{x\rightarrow 0}\left [ \left ( \dfrac{1}{2}-\dfrac{1}{3} \right )+ term\  containing\  x \right ]=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$
(Alternatively apply L-Hospital's Rule)

Let $\displaystyle f(x)=\left ( 2-\dfrac{x}{a} \right )^{\tan \left ( \dfrac{\pi :x }{2:a} \right )}, x\neq a$. The value which should be assigned to $f$ at $x=a$ so that it is continuous everywhere is

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $\dfrac{2}{\pi} $

  2. $e^{-2/\pi }$

  3. $2$

  4. $e^{2/\pi }$

Show answer
Correct Option: D
Explanation:

For $f$ to be continuous, we must have
$\displaystyle f(a)=\lim _{x\rightarrow a}\left ( 2-\dfrac{x}{a} \right )^{\tan \left ( \dfrac{\pi :x }{2:a} \right )}$
$\Rightarrow log  f(x)=\tan \left ( \dfrac{\pi :x }{2:a} \right )log(2-\dfrac{x}{a})$
$=\displaystyle \dfrac{log(2-\dfrac{x}{a})}{\cot \left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )}\left ( \dfrac{0}{0}form \right )$
$\displaystyle \lim _{x\rightarrow a} log f(x)=\lim _{x\rightarrow a}\dfrac{\dfrac{-1/a}{2-\dfrac{x}{a}}}{-\dfrac{\pi }{2a}\csc ^{2}\left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )}$
(L'Hospital's Rule)
$=\displaystyle \dfrac{2}{\pi }\lim _{x\rightarrow a}\dfrac{1}{2-\dfrac{x}{a}}\times \lim _{x\rightarrow a}\sin ^{2}\left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )$
$=\displaystyle \dfrac{2}{\pi }$
Therefore, $\displaystyle \lim _{x\rightarrow a}f(x)=e^{2/\pi }$
Hence, option 'D' is correct.

If $f(x)=\left{\begin{matrix} |x|-3, & x < 1\ |x-2|+a, & x\geq 1\end{matrix}\right.$ and $g(x)=\left{\begin{matrix} 2-|x|, & x < 2 \ sgn(x)-b, & x\geq 2\end{matrix}\right.$ and $h(x)=f(x)+g(x)$ is discontinuous at exactly one point, then which of the following values of a and b are possible.

mathematics and statistics continuity discontinuity and its types types of discontinuity differencial calculus - limits and continuity

  1. $a=-3, b=0$

  2. $a=2, b=1$

  3. $a=2, b=0$

  4. $a=-3, b=1$

Show answer
Correct Option: A,B