Question

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.

Answer 1

$$\sqrt{1+x}=(1+x)^{1/2}$$

$$=1+\dfrac{1}{2}x+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)}{2!}x^2+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)(\dfrac{1}{2}-2)}{3!}x^3+...$$

$$f(x)=x\sqrt{1+x}=x+\dfrac{1}{2}x^2-\dfrac{1}{8}x^3+\dfrac{1}{16}x^3-\dfrac{5}{128}x^4+...$$

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

By the calculator

$$g(x)=x(\cosh x)^{1/x}\approx x+\dfrac{1}{2}x^2+\dfrac{1}{8}x^3-\dfrac{1}{16}x^4-\dfrac{5}{128}x^5+...$$

$$x\sqrt{1+x}+x(\cosh x)^{1/x}\approx2x+x^2-\dfrac{5}{64}x^5$$

$$e^x-1=x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+\dfrac{1}{120}x^5+...$$

$$(e^x-1)^3\approx x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5$$

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

$$=\lim\limits_{x\to0}(\dfrac{4(2x+x^2-\dfrac{5}{64}x^5)}{x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5})$$

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