Solving functions? Ouch!

Calculus Level pending

If f ( x ) = x 2 g ( x ) f(x)=x^2g(x) and lim ⁡ x → 0 g ( x ) c o s e c ( x ) = f ′ ′ ′ ( 0 ) / λ \lim_{x \to 0} g(x)cosec(x)=f'''(0)/λ .The value of λ is?


The answer is 6.

Sanchayan Dutta by Sanchayan Dutta , 23, India

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1 solution

Tom Engelsman
May 13, 2021

The function we are interested in computing the limit for can be rewritten as g ( x ) ⋅ csc ⁡ ( x ) = f ( x ) x 2 sin ⁡ ( x ) = f ( x ) h ( x ) \large g(x) \cdot \csc(x) = \frac{f(x)}{x^2 \sin(x)} = \frac{f(x)}{h(x)} . The limit requires we apply L'Hopital's Rule thrice since we have f ′ ′ ′ ( 0 ) f'''(0) in the numerator. The third derivative of the denominator computes to:

h ′ ′ ′ ( x ) = ( 6 − x 2 ) cos ⁡ ( x ) − 6 x sin ⁡ ( x ) ⇒ λ = h ′ ′ ′ ( 0 ) = ( 6 − 0 ) cos ⁡ ( 0 ) − 6 ( 0 ) sin ⁡ ( 0 ) = 6 . h'''(x) = (6-x^2)\cos(x) -6x \sin(x) \Rightarrow \lambda = h'''(0) = (6-0)\cos(0) -6(0)\sin(0) = \boxed{6}.

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