There's a quartic root in my limit

Calculus Level 4

lim ⁡ x → 0 1 − x 2 − 1 − 4 x 2 + x 4 4 x 4 = ? \lim_{x\to 0 } \dfrac{ 1 -x^2 -\sqrt[4]{1-4x^2 + x^4}}{x^4} = \, ?


The answer is 1.25.

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Andrea Virgillito by Andrea Virgillito , 22, Italy

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1 solution

Andrea Virgillito
Feb 14, 2017

We have that: 1 − 4 x 2 + x 4 4 = 1 + 1 4 ( − 4 x 2 + x 4 ) + 1 4 ( 1 4 − 1 ) 2 ! ( − 4 x 2 + x 4 ) 2 + 1 4 ( 1 4 − 1 ) ( 1 4 − 2 ) 3 ! ( − 4 x 2 + x 4 ) 3 + 1 4 ( 1 4 − 1 ) ( 1 4 − 2 ) ( 1 4 − 3 ) 4 ! ( − 4 x 2 + x 4 ) 4 + o ( x 4 ) = 1 − x 2 − 5 4 x 4 + o ( x 4 ) \sqrt[4]{1-4x^2+x^4}= 1+\frac{1}{4}(-4x^2+x^4)+\frac{\frac{1}{4}(\frac{1}{4}-1)}{2!}(-4x^2+x^4)^2+\frac{\frac{1}{4}(\frac{1}{4}-1)(\frac{1}{4}-2)}{3!}(-4x^2+x^4)^3+ \frac{\frac{1}{4}(\frac{1}{4}-1)(\frac{1}{4}-2)(\frac{1}{4}-3)}{4!}(-4x^2+x^4)^4+o(x^4)= 1-x^2-\frac{5}{4}x^4+o(x^4)

Thus: lim ⁡ x → 0 1 − x 2 − 1 − 4 x 2 + x 4 4 x 4 = lim ⁡ x → 0 1 − x 2 − ( 1 − x 2 − 5 4 x 4 ) + o ( x 4 ) x 4 = 5 4 = 1.25 \lim_{x\to 0 } \dfrac{ 1 -x^2 -\sqrt[4]{1-4x^2 + x^4}}{x^4} = \lim_{x\to 0 } \dfrac{ 1 -x^2 -(1-x^2-\frac{5}{4}x^4)+o(x^4)}{x^4} =\frac{5}{4}=1.25

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Not an elegant method.

沂泓 纪 - 4 years, 3 months ago

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I know there are some methods more elegant but I have chosen to write that just because I really like Taylor :)

Andrea Virgillito - 4 years, 3 months ago

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Lol,I see.

沂泓 纪 - 4 years, 3 months ago

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