If N = n!(n n ∈ N, n > 2, then find
limN→∞log2N- 1 + log3N- 1 + ... + lognN- 1
Use the formula limx→0ax - 1x = logea to compute limx→02x - 11 + x - 1
The value of limx→0sinx + cosx - 1x2
1
12
- 12
0
B.
limx→0sin2x + cosx - 1x2= limx→0cosx - cos2xx2= limx→01 - cosxx2= limx→02sin2x2x2cosx= limx→02sin2x2x22 . 4 . limx→0cosx= 24 × 1 = 12
The value of limx→01 + 5x21 + 3x21x2
e2
e
1e
1e2
If f(5)=7 and f'(5)=7 then limx→5xf(5) - 5fxx - 5 is given by
35
- 35
28
- 28
The value of f(0) so that the function 1 - cos1 - cosxx4 is continuous everywhere is
14
16
18
limx→0sinxx is equal to
positive infinity
does not exist
If y = tan-11 - sinx1 + sinx, then the value of dydx at x = π6 is
- 1
The value of the limit limx→1sinex - 1 - 1logx
Let f(x) = x + 3x + 1, then the value of limx→- 3 - 0fx is