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Find the value of a for which the limit of function exists at x = 0. f(x) = (1-cos4x)/x 2 , x f(x) = a, x = 0 f(x) = (x) 1/2 /((16 + x 1/2 ) 1/2 – 4), x > 0

Find the value of a for which the limit of function exists at x = 0.
f(x) = (1-cos4x)/x2, x
f(x) = a, x = 0
f(x) = (x)1/2/((16 + x1/2)1/2 – 4), x > 0

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Since the limit of function exist at zero. So at x = 0
LHL = f(0) = RHL
LHL = \lim_{x\rightarrow 0}\frac{1-cos4x}{x^{2}}
Apply L’Hospital rule
LHL = \lim_{x\rightarrow 0}\frac{4sin4x}{2x}
LHL = \lim_{x\rightarrow 0}\frac{2sin4x}{x}
LHL = \lim_{x\rightarrow 0}\frac{2.4cos4x}{1} = 8
LHL = a
a = 8

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