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Multiply the Maclaurin series for the exponential and the sine, keeping terms through degree three:Only $x$, $x^2$, and $(1/2-1/6)x^3$ contribute at these orders, so

Near zero the logarithm's argument is positive, so $\ln [(2+x{)}^3]=3\ln (2+x)=3\ln 2+3\ln (1+x/2)$. Use the Maclaurin series of the natural logarithm:The local calculation uses $x>-2$ for the real logarithm; the power series converges for $|x|<2$.

Insert $u=-1/x^2$ into the exponential series $e^u={\sum }_{n=0}^{\infty }{u}^n/n!$. For every fixed $x\ne 0$, this gives the expansion with absolute convergenceThis is a Laurent series in inverse powers of $x$, as requested. It is not a Maclaurin series about zero: that series has every coefficient zero and cannot equal $f$ on a punctured neighborhood of zero. There is no contradiction, since the inverse-power expansion does not apply at $x=0$.

Use the Maclaurin series on the positive side:In the last expression the original square root covers $e^x-1$, whose leading term is $x>0$. Hence $g/f\to 0$ and $f/h\to 0$. All three are positive for sufficiently small positive $x$, soThus $h$ is greatest and $g$ least in the stated limit.

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

Multiply the Maclaurin series $e^{-x}=1-x+x^2/2+O(x^3)$ by the binomial series $(1+x{)}^{-1/2}=1-x/2+3x^2/8+O(x^3)$. The constant, linear and quadratic coefficients are therefore

Use the Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $(1+x{)}^{-2}=1-2x+3x^2+O(x^3)$. Multiplying through the cubic order givesThese are the first three nonzero terms; the constant term vanishes.

Use $\cosh x=1+x^2/2+x^4/24+x^6/720+O(x^8)$. The Maclaurin series $\ln (1+u)=u-u^2/2+u^3/3+O(u^4)$ givesAll three nonzero terms are needed; stopping at fourth order would supply only two.

Expand the natural logarithm and reciprocal separately:Multiplication gives the first three nonzero terms of the Maclaurin series:The local expansions converge for $|x|<2$.

The Maclaurin series give $e^{-x^3}=1-x^3+O(x^6)$ and $\cosh x=1+x^2/2+x^4/24+O(x^6)$. Inverting the latter using the geometric series givesMultiplying the two expansions yieldsThe remainder is $O(x^5)$ because the product already generates an $x^5$ term from $(-x^3)(-x^2/2)$.

Use the Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $e^u=1+u+u^2/2+u^3/6+u^4/24+O(u^5)$. The terms through fourth degree giveIn particular the cubic coefficient cancels; it has not been omitted accidentally.

Multiply the Maclaurin series of hyperbolic cosine by the binomial series for the reciprocal square root:The quadratic terms cancel, while the quartic coefficient is $1/24-1/4+3/8=1/6$. HenceThe absence of odd powers follows from the function's even symmetry.

Multiply the Maclaurin series $\ln (1+x^2)=x^2-x^4/2+O(x^6)$ and the geometric series $(1+x^2{)}^{-1}=1-x^2+O(x^4)$. The first two nonzero terms are

Use the Maclaurin series of the logarithm and hyperbolic cosine:To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. ThusThe numerator in the original PDF is $\ln (1+x^3)$; the cubic power is missing from the local TeX transcription.

Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

Divide numerator and denominator by $x$ before taking their Maclaurin series:For the reciprocal denominator, the geometric series gives $1-x^2/6+7x^4/360+O(x^6)$. Multiplying gives the Taylor expansionStrictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

The Maclaurin series of the hyperbolic cosine and cosine areBoth are even, so write their quotient as $1+a{x}^2+b{x}^4+O(x^6)$. Multiplication by the denominator gives coefficients $a-1/2$ and $b-a/2+1/24$. Matching the numerator yields $a=1$ and $b=1/2$. ThusThe denominator is nonzero near the expansion point, so division of the power series is valid there.

Use the Maclaurin series $\sin 3x=3x-\frac{9}{2}{x}^3+O(x^5)$ and $\cos x=1-\frac{1}{2}{x}^2+O(x^4)$. Multiplication givesThe same terms follow from the two sine harmonics in the Fourier series found above.

The Maclaurin series $\sinh t=t+t^3/3!+t^5/5!+O(t^7)$, applied with $t=x^2$ and then multiplied by $x$, givesThese are the first three nonzero terms. The remainder is written in Big O notation.

Factor the leading term out of the Maclaurin series for the sine:The binomial theorem gives, through degree four inside the bracket,The coefficient of $x^4$ is $1/20+5/12=7/15$. Multiplying back by $x^6$ yields