29Ans 2....
sinx < x for any value of x..so in the first one sinx/x < 1 and [something < 1] = 0..
in the second part we know that limx→0 sinx/x = 1..so [1] = 1
correct me if i am wrong
Q1 Let h(x) = l kx+5 l and domain of f(x) and f(h(x)) are [-5,7] and [-6,1] respectively, then value of k is _____ (the answer is X if cant say)
Q2. lim(x->0) [sinxx] = _____
[lim(x-->0) sinxx] = ______
-
UP 0 DOWN 0 0 7
7 Answers
29
106but still the limit in the second case is NOT 1.
It is 1- .. So, [1-] = 0
naa??
11asish u said
but still the limit in the second case is NOT 1.
It is 1- .. So, [1-] = 0
naa??
bro ,if we have 0- then both the numerator and denominator will be 0 coz sin (-x)=-sinx