Select Board & Class
E.g: 9876543210, 01112345678
We will give you a call shortly, Thank You
Office hours: 9:00 am to 9:00 pm IST (7 days a week)
Syllabus
(a) (–1)k (k – 1)π (b) (–1)k – 1 (k – 1)π (c) (–1)k kπ (d) (–1)k – 1 kπ
1. g is not defined at x=0
2. g is not continuous at x=0
3. The limit of g(x) as x approaches 0 equals 1
4. g'(0)=1
I don't know why it is a function as the graph is not continuous. For the value of x to be π/2 i.e., tan π/2, the value of the y goes to positive infinity from left-hand side and the negative infinity from the right-hand side. Moreover, there is asymptote at x = π/2, and thus even there does not exist any limit.
Q).
1. 0
2. -2/3
3 3
4. 1
question no.8
1.
2.
3.
4.
(a) 0
(b)
(c)
(d) none of these
1. -1
2. 1
3. 2
4. -2
Q. Determine the intervals on which the given function is continuous:
1.
2.
3.
4.
A. lim h tends to 0 k-h-[k-h]
= what to do next sir
in book they are doing like this
= lim htends to 0 k-h-(k-1) = 1
how are they puting 1 over ther ? plz explain.