find the value of k such thatf(x) = 1-cos4x/8x^2,if x is not equal to zero                     k,if x=0it is continous at x=0

Nishant Vora IIT Patna
9 years ago
Hi Student,

For the fucntion to be continuous at x=0, f(0+)=f(0-)=f(0)=k

So, Lim(x->0) [1-cos4x/8x^2]
=> Lim(x->0) [2sin22x/8x^2]
=> Lim(x->0) [sin22x/(2x)2]

=> 1

Rasal V
10 Points
9 years ago
K=1.
lim      f(x)=k
X-->0-

therefore, lim              1-cos4x/8x^2=sin^2(2x)+cos^2(2x)-cos^2(2x)+sin^2(2x)/8x^2
x-->0
=1/8 lim  2sinj^2(2x)/x^2
x-->0
=1/8*2*lim      4    since(lim x-->0 sinx/x=1)
x-->0

therefore k=1/8*2*4=1
Rasal V
10 Points
9 years ago
K=1.
lim      f(x)=k
X-->0-

therefore, lim              1-cos4x/8x^2=sin^2(2x)+cos^2(2x)-cos^2(2x)+sin^2(2x)/8x^2
x-->0
=1/8 lim  2sinj^2(2x)/x^2
x-->0
=1/8*2*lim      4    since(lim x-->0 sinx/x=1)
x-->0

therefore k=1/8*2*4=1
Rasal V
10 Points
9 years ago
K=1.
lim      f(x)=k
X-->0-

therefore, lim              1-cos4x/8x^2=sin^2(2x)+cos^2(2x)-cos^2(2x)+sin^2(2x)/8x^2
x-->0
=1/8 lim  2sinj^2(2x)/x^2
x-->0
=1/8*2*lim      4    since(lim x-->0 sinx/x=1)
x-->0

therefore k=1/8*2*4=1