Answer to Question #156276 in Calculus for Eva

Question #156276
Write down the Maclaurin expression of the following functions up to and including the term x^4
f:x → f(x) = x sqrt(1 + x) and g:x → g(x) = x(coshx)^(1/x). Hence show that
Lim as x→0 [ (4(x*sqrt(1+x) - x*(coshx)^(1/x)) / (e^x - 1)^3 ] = -1 with x ≠ 0.
1
Expert's answer
2021-01-29T05:22:23-0500
1+x=(1+x)1/2\sqrt{1+x}=(1+x)^{1/2}

=1+12x+12(121)2!x2+12(121)(122)3!x3+...=1+\dfrac{1}{2}x+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)}{2!}x^2+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)(\dfrac{1}{2}-2)}{3!}x^3+...

f(x)=x1+x=x+12x218x3+116x35128x4+...f(x)=x\sqrt{1+x}=x+\dfrac{1}{2}x^2-\dfrac{1}{8}x^3+\dfrac{1}{16}x^3-\dfrac{5}{128}x^4+...

The function g(x)=x(coshx)1/xg(x)=x(\cosh x)^{1/x} is not defined at x=0.x=0. Therefore we cannot write down the Maclaurin expression of the following function.

By the calculator


g(x)=x(coshx)1/xx+12x2+18x3116x45128x5+...g(x)=x(\cosh x)^{1/x}\approx x+\dfrac{1}{2}x^2+\dfrac{1}{8}x^3-\dfrac{1}{16}x^4-\dfrac{5}{128}x^5+...

x1+x+x(coshx)1/x2x+x2564x5x\sqrt{1+x}+x(\cosh x)^{1/x}\approx2x+x^2-\dfrac{5}{64}x^5

ex1=x+12x2+16x3+124x4+1120x5+...e^x-1=x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+\dfrac{1}{120}x^5+...

(ex1)3x3+32x4+54x5(e^x-1)^3\approx x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5

limx0(4(x1+x+x(coshx)1/x)(ex1)3)\lim\limits_{x\to0}(\dfrac{4(x\sqrt{1+x}+x(\cosh x)^{1/x})}{(e^x-1)^3})

=limx0(4(2x+x2564x5)x3+32x4+54x5)=\lim\limits_{x\to0}(\dfrac{4(2x+x^2-\dfrac{5}{64}x^5)}{x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5})

=14,x0=-\dfrac{1}{4},x\not=0



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