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Differential Calculus> Find the value of a for which the limit o...

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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 + x^1/2)^1/2 – 4), x > 0

Lovey , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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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