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A continuous function can have some points where limit does not exist. - Mathematics

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प्रश्न

A continuous function can have some points where limit does not exist.

विकल्प

  • True

  • False

MCQ
सत्य या असत्य

उत्तर

This statement is False.

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अध्याय 5: Continuity And Differentiability - Solved Examples [पृष्ठ १०६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 5 Continuity And Differentiability
Solved Examples | Q 44 | पृष्ठ १०६

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

Examine the continuity of the following function :

`{:(,,f(x)= x^2 -x+9,"for",x≤3),(,,=4x+3,"for",x>3):}}"at "x=3`


Examine the following function for continuity:

`f (x)1/(x - 5), x != 5`


Examine the following function for continuity:

f(x) = | x – 5|


Show that

\[f\left( x \right)\] = \begin{cases}\frac{x - \left| x \right|}{2}, when & x \neq 0 \\ 2 , when & x = 0\end{cases}

is discontinuous at x = 0.

 

Discuss the continuity of the following functions at the indicated point(s): 

\[f\left( x \right) = \left\{ \begin{array}{l}\frac{1 - x^n}{1 - x}, & x \neq 1 \\ n - 1 , & x = 1\end{array}n \in N \right.at x = 1\]

Discuss the continuity of the function f(x) at the point x = 1/2, where \[f\left( x \right) = \begin{cases}x, 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, x = \frac{1}{2} \\ 1 - x, \frac{1}{2} < x \leq 1\end{cases}\] 


In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}k( x^2 + 2), \text{if} & x \leq 0 \\ 3x + 1 , \text{if} & x > 0\end{cases}\]


The function  \[f\left( x \right) = \begin{cases}\frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \neq 0 \\ 0 , & x = 0\end{cases}\]

 


If the function \[f\left( x \right) = \begin{cases}\left( \cos x \right)^{1/x} , & x \neq 0 \\ k , & x = 0\end{cases}\] is continuous at x = 0, then the value of k is


The value of f (0) so that the function 

\[f\left( x \right) = \frac{2 - \left( 256 - 7x \right)^{1/8}}{\left( 5x + 32 \right)^{1/5} - 2},\]  0 is continuous everywhere, is given by


The value of k which makes \[f\left( x \right) = \begin{cases}\sin\frac{1}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\]    continuous at x = 0, is

 


The points of discontinuity of the function 

\[f\left( x \right) = \begin{cases}2\sqrt{x} , & 0 \leq x \leq 1 \\ 4 - 2x , & 1 < x < \frac{5}{2} \\ 2x - 7 , & \frac{5}{2} \leq x \leq 4\end{cases}\text{ is } \left( \text{ are }\right)\] 


If  \[f\left( x \right) = \begin{cases}\frac{\sin \left( \cos x \right) - \cos x}{\left( \pi - 2x \right)^2}, & x \neq \frac{\pi}{2} \\ k , & x = \frac{\pi}{2}\end{cases}\]is continuous at x = π/2, then k is equal to


Is every continuous function differentiable?


The set of points where the function f (x) = x |x| is differentiable is 

 


The function f (x) = e|x| is


If \[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,\] 

then at x = 0, f (x)


If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]


The function f (x) =  |cos x| is


Discuss continuity of f(x) =`(x^3-64)/(sqrt(x^2+9)-5)` For x ≠ 4 

= 10 for x = 4  at x = 4


Discuss the continuity of f at x = 1
Where f(X) = `[ 3 - sqrt ( 2x + 7 ) / ( x - 1 )]`           For x ≠ 1
                    = `-1/3`                                                 For x = 1


If the function
f(x) = x2 + ax + b,         x < 2

      = 3x + 2,                 2≤ x ≤ 4

      = 2ax + 5b,             4 < x

is continuous at x = 2 and x = 4, then find the values of a and b


Show that the function f defined by f(x) = `{{:(x sin  1/x",", x ≠ 0),(0",", x = 0):}` is continuous at x = 0.


Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if"  x ≠ 0),(0",",  "if"  x = 0):}` is discontinuous at x = 0.


The number of points at which the function f(x) = `1/(x - [x])` is not continuous is ______.


The function given by f (x) = tanx is discontinuous on the set ______.


The function f(x) = |x| + |x – 1| is ______.


f(x) = `{{:((1 - cos 2x)/x^2",", "if"  x ≠ 0),(5",", "if"  x = 0):}` at x = 0


f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1


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