Let's do some calculus! (29)

Calculus Level 4

lim ⁡ x → 0 sin ⁡ x 4 − x 4 cos ⁡ x 4 + x 20 x 4 ( e 2 x 4 − 2 x 4 − 1 ) = ? \large \displaystyle \lim_{x \to 0} \dfrac{\sin x^4 - x^4 \cos x^4 + x^{20}}{x^4 \left( e^{2{x}^{4}} - 2 x^4 - 1 \right)} ~ = ~ ?

Submit your answer rounded up to 3 places of decimal.

Notation: e ≈ 2.71828 e \approx 2.71828 is the Euler's number .

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The answer is 0.167.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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2 solutions

Chew-Seong Cheong
Sep 25, 2016

L = lim ⁡ x → 0 sin ⁡ x 4 − x 4 cos ⁡ x 4 + x 20 x 4 ( e 2 x 4 − 2 x 4 − 1 ) Using Maclaurin series = lim ⁡ x → 0 x 4 − x 12 3 ! + x 20 5 ! − . . . − x 4 ( 1 − x 8 2 ! + x 16 4 ! − . . . ) + x 20 x 4 ( 1 + 2 x 4 1 ! + 4 x 8 2 ! − . . . − 2 x 4 − 1 ) = lim ⁡ x → 0 ( 1 2 ! − 1 3 ! ) x 12 − ( 1 4 ! − 1 5 ! + 1 ) x 20 + O ( x 24 ) 4 x 12 2 ! − 8 x 16 4 ! + O ( x 20 ) Divide up and down by x 12 = lim ⁡ x → 0 1 3 − 31 x 8 30 + O ( x 12 ) 2 − x 16 3 + O ( x 8 ) = 1 6 ≈ 0.167 \begin{aligned} L & = \lim_{x \to 0} \frac {\color{#3D99F6}{\sin x^4}-x^4\color{#D61F06}{\cos x^4}+x^{20}}{x^4(\color{#20A900}{e^{2x^4}}-2x^4-1)} & \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \frac {\color{#3D99F6}{x^4-\frac {x^{12}}{3!}+ \frac {x^{20}}{5!}-...}-x^4\left(\color{#D61F06}{1-\frac {x^8}{2!}+ \frac {x^{16}}{4!}-...}\right)+x^{20}}{x^4(\color{#20A900}{1+\frac {2x^4}{1!}+ \frac {4x^8}{2!}-...}-2x^4-1)} \\ & = \lim_{x \to 0} \frac {\left(\frac 1{2!} - \frac 1{3!} \right)x^{12} - \left(\frac 1{4!} - \frac 1{5!} + 1\right)x^{20} + O(x^{24})}{\frac {4x^{12}}{2!}-\frac {8x^{16}}{4!}+O(x^{20})} & \small \color{#3D99F6}{\text{Divide up and down by }x^{12}} \\ & = \lim_{x \to 0} \frac {\frac 13 - \frac {31x^8}{30} + O(x^{12})}{2-\frac {x^{16}}{3}+O(x^{8})} \\ & = \frac 16 \approx \boxed{0.167} \end{aligned}

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I don't really know the application of the 'big O O ( ⋅ ) O ( \cdot) ' notation that you've used in your solution. I've seen them on Wikipedia but don't know what are they.

Tapas Mazumdar - 4 years, 8 months ago

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In this case, O ( x 12 ) = a 3 x 12 + a 4 x 16 + a 5 x 20 + . . . O(x^{12}) = a_3x^{12} + a_4x^{16} + a_5x^{20}+... .

Chew-Seong Cheong - 4 years, 8 months ago

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Can you refer me some good link for me to understand this?

Tapas Mazumdar - 4 years, 8 months ago

We can Put t = x 4 t=x^4 to reduce calculations

Sabhrant Sachan - 4 years, 8 months ago

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Yes, I should have seen that.

Chew-Seong Cheong - 4 years, 8 months ago

Really tedious solution using L'Hôpital's rule brute force

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