4@Nishant Sir :
1) Lt (x-->0) xsin(1/x) = sin(1/x)/(1/x) = 1 ??? ≠0
Then can v say it's continuous at x=0
2) it's discontinuous
3)same doubt as in the case of (!) [7]
4)Ya i got it It's continuous at origin
Which of the following fcns defined below are continuous at the origin ?
(A) f(x) = x sin(1/x) if x ≠0
0 if x=0
(B) g(x) = [cos(x2-5x+6)] / (x2 - 5x +6) if x≠2,3
1 if x=2,3
(C) h(x) = xtan-1(1/x) if x≠0
0 if x=0
(D) p(x) = sin(x+1)/(x+1) if x≠0
1 if x=0
-
UP 0 DOWN 0 1 4
4 Answers
62first one is continuous at x=0 since x sin x is less than equal to x and greater than equal to -x.. so near zero it will be exactly zero..
Hence it is continuous
The second one is cos t/ t when t is close to zero.. which will give us a form of 1 divided by zero.. which will not be defined..
The 3rd part the limit is again zero.. That is because tan-1 is between -pi/2 and pi/2
so when very close to zero.. the limit of h(x) will be zero.. so it will be continuous..
Now can you solve the last one?
4
21lim x=> 0 sin(1/x)/(1/x) not equal to 1
if f(x)→0 wen x→a
then only lim x→a sinf(x)/f(x)=1
here in this case f(x)=1/x and x=>0 but f(x) i.e 1/x is =>∞
hence lim x=> 0 sin(1/x)/(1/x) not equal to 1
4oops mistake again
Thanks Deepak for correcting
&Thanks Nishant Sir I got it now