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Which one of the following functions is continuous at $x = 3?$

  1. $f(x) = \begin{cases} 2,&\text{if $x = 3$ } \\ x-1& \text{if $x > 3$}\\ \frac{x+3}{3}&\text{if $x < 3$ } \end{cases}$
  2. $f(x) = \begin{cases} 4,&\text{if $x = 3$ } \\ 8-x& \text{if $x \neq 3$} \end{cases}$
  3. $f(x) = \begin{cases} x+3,&\text{if $x \leq 3$ } \\ x-4& \text{if $x > 3$} \end{cases}$
  4. $f(x) = \begin{cases} \frac{1}{x^3-27}&\text{if $x \neq 3$ } \end{cases}$
in Calculus 3.9k views
1
is it's ans be a?
3
$Properties\ of\ Discontinuity\\ A\ function\ f(x)\ will\ be\ discontinuous\ at\ x=a\ here\ x=3, if\ any\ one\ below\ hold\\ \\ 1. \lim_{x->a^{-}}f(x) \neq \lim_{x->a^{+}}f(x)\ (LHL\neq RHL)\\ 2. \lim_{x->a^{-}}f(x)\doteq \lim_{x->a^{+}}f(x)\neq f(a)\\ 3. f(a)\ is\ not\ defined\\ 4. At\ least\ one\ of\ the\ limit\ not\ exists$

3 Answers

36 votes
 
Best answer
For continuity, Left hand limit must be equal to right hand limit. For continuity at $x = 3$,

the value of $f(x)$ just above and just below $3$ must be the same.

A. $f(3) = 2. f(3+) = x - 1 = 2. f(3-) =\frac{(x+3)}{3} =\frac{6}{3} = 2.\text{ Hence continuous.}$

B. $f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. \text{ So, not continuous.}$

C. $f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1.\text{ So, not continuous.}$

D. $f(3)\text{ is not existing. So, not continuous.}$
Correct Answer: $A$

edited by
0
@Arjun Sir, why D is not existing? plzz explain ...am little bit poor in Maths so plzzz
2
at x=3 D is approaching infinity and hence function is not defined at x=3. For continuity LHL=RHL=value of the function. Hence not continuous.
1
In option B, Limit exists since $\lim_{x \to 3^{-}}{f(x)} = \lim_{x \to 3^{+}}{f(x)}$, but $ \lim_{x \to 3}{f(x)}\neq f(3)$
1

yes option D is a discontinuous  graph 

5 votes

So this f(x) is continuous at x=3

1 vote
option A is true because LHL = RHL

LHL=$\lim_{x->3}(x+x)/3=(3+3)/3=2$

RHL=$\lim_{x->3}(x-1)=(3-1)=2$
0
LHL=limx->3-(x+3)/3 not (x+x)/3.
Answer:

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