find k if f(x) is continous at x=0.
f(x)={(1-cos4x)/x^2 if x k if x=0
(√x)/√16+root of x-4 }

Consider the following function.   fx=1-cos4xx2, when x<0           =k,  x=0           =x16+x-4 , when x>0  LHL =  limx0-fx=limx0-1-cos4xx2 put x = 0-h, as x0-, then h0LHL = limh01 - cos 4hh2 =limh02 sin22hh2 = 2 limh0sin 2h2h×2limh0sin 2h2h×2 = 8

  RHL =limx0+fx=   limx0+x16+x-4  =limh0 h16+h - 4  put x = 0+h, as x0+, then h0=limh0 h16+h - 4×16+h + 416+h + 4=limh0 h16+h + 416+h - 16=limh0 16+h + 4 = 8    Also, f0 = kSince f is continuous at x = 0, thenLHL = RHL = f08 = 8 = kk = 8                          

  • 46
for f(x) to be continuous at x=0

lim _ f (x)=f (0)
x->0
lim _ f (x)=lim (1-cos 4x)/x^(2)
x->0 x->0

=lim 2 sin^(2) 2x/x^(2)
x->0

=lim (2*4) sin^(2) 2x/ (2^(2) * x^(2))
x->0

=8lim (sin 2x/2x)^(2)
x->0

=8*1 [using lim sin x/x =1]
x->0

=8
=f (0)
=k
Hence, k=8
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