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question mark

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

Rohan joseph , 11 Years ago
Grade 11
anser 4 Answers
Nishant Vora
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

Last Activity: 11 Years ago
Rasal V
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
Last Activity: 11 Years ago
Rasal V
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
Last Activity: 11 Years ago
Rasal V
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
Last Activity: 11 Years ago
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