17 votes 17 votes Which one of the following functions is continuous at $x = 3?$ $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}$ $f(x) = \begin{cases} 4,&\text{if $x = 3$ } \\ 8-x& \text{if $x \neq 3$} \end{cases}$ $f(x) = \begin{cases} x+3,&\text{if $x \leq 3$ } \\ x-4& \text{if $x > 3$} \end{cases}$ $f(x) = \begin{cases} \frac{1}{x^3-27}&\text{if $x \neq 3$ } \end{cases}$ Calculus gatecse-2013 calculus continuity normal + – Arjun asked Sep 24, 2014 Arjun 7.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply amkrj commented Apr 9, 2015 reply Follow Share is it's ans be a? 1 votes 1 votes manish_dt commented Jul 14, 2020 reply Follow Share $Properties\ of\ Discontinuity\\ A\ function\ f(x)\ will\ be\ discontinuous\ at\ x=a\ here\ x=3, if\ any\ one\ below\ hold\\ \\ 1. \lim_{x->a^{-}}f(x) \neq \lim_{x->a^{+}}f(x)\ (LHL\neq RHL)\\ 2. \lim_{x->a^{-}}f(x)\doteq \lim_{x->a^{+}}f(x)\neq f(a)\\ 3. f(a)\ is\ not\ defined\\ 4. At\ least\ one\ of\ the\ limit\ not\ exists$ 4 votes 4 votes Please log in or register to add a comment.
Best answer 46 votes 46 votes 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. $f(3) = 2. f(3+) = x - 1 = 2. f(3-) =\frac{(x+3)}{3} =\frac{6}{3} = 2.\text{ Hence, continuous.}$ $f(3) = 4. f(3+) = f(3-) = 8 - 3 = 5. \text{ So, not continuous.}$ $f(3) = f(3-) = x + 3 = 6. f(3+) = x - 4 = -1.\text{ So, not continuous.}$ $f(3)\text{ is not existing. So, not continuous.}$ Correct Answer: $A$ Arjun answered Apr 9, 2015 • edited Mar 27, 2021 by soujanyareddy13 Arjun comment Share Follow See all 4 Comments See all 4 4 Comments reply aman.anand commented Jan 2, 2017 reply Follow Share @Arjun Sir, why D is not existing? plzz explain ...am little bit poor in Maths so plzzz 0 votes 0 votes sushmita commented Jan 8, 2017 reply Follow Share 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. 2 votes 2 votes Pranav Kant Gaur commented Jan 30, 2017 reply Follow Share 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 votes 1 votes Prateek kumar commented Jan 14, 2018 reply Follow Share yes option D is a discontinuous graph 3 votes 3 votes Please log in or register to add a comment.
6 votes 6 votes So this f(x) is continuous at x=3 VIKAS TIWARI answered Jan 5, 2018 VIKAS TIWARI comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes option A is true because LHL = RHL LHL=$\lim_{x->3}(x+x)/3=(3+3)/3=2$ RHL=$\lim_{x->3}(x-1)=(3-1)=2$ himanshukumarpatel answered Apr 10, 2018 himanshukumarpatel comment Share Follow See 1 comment See all 1 1 comment reply Sona Barman commented Dec 11, 2020 reply Follow Share LHL=limx->3-(x+3)/3 not (x+x)/3. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes . akshay_123 answered Sep 8, 2023 akshay_123 comment Share Follow See all 0 reply Please log in or register to add a comment.