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Use the Maclaurin series $\ln (1+x)=x-x^2/2+x^3/3+O(x^4)$ and the geometric series $(1+y{)}^{-1}=1-y+y^2-y^3+\cdots$. Multiplication, grouped by total degree, givesThe six displayed nonzero terms include every contribution through total degree three. A sufficient open region for the convergent product expansion is $|x|<1$, $|y|<1$.

The Maclaurin series $\sin x=x-x^3/6+O(x^5)$ cancels the numerator's linear term, leavingThus $\boxed{\underset{x\to 0}{\lim }(\sin x-x)/x^3=-1/6}$. The same expansion applies for positive and negative $x$, so this is a two-sided limit.

Combine the terms before taking their limit:Using the exponential function Maclaurin series, the numerator is $x^2/2+O(x^3)$ and the denominator is $x^2+O(x^3)$. ConsequentlyThe cancellation removes the two individual singularities. Subtracting their separate divergent limits would not determine the finite answer.

Write $\tan x=a_{1}x+a_3{x}^3+a_5{x}^5+O(x^7)$. Multiplying by the cosine Maclaurin series and comparing with the sine series givesCoefficient comparison yields $a_1=1$, $a_3-a_1/2=-1/6$ and $a_5-a_3/2+a_1/24=1/120$. HenceSubstitute $u=\tan x$ into $e^u=1+u+u^2/2+u^3/6+O(u^4)$. Keeping one extra order to make the term-counting explicit,Counting the constant term, the first three nonzero terms are $\boxed{1+x+x^2/2}$. If the phrase is taken to mean the first three nonconstant terms, they are $x+x^2/2+x^3/2$; the displayed expansion supplies both conventions. The cubic coefficient is $1/3+1/6=1/2$, combining the cubic term in $\tan x$ with the cubic exponential term.

Start with the Maclaurin series $\cos x=1-x^2/2+x^4/24+O(x^6)$. Put $z=-x^2/2+x^4/24+O(x^6)$ into the binomial series $(1+z{)}^{-2}=1-2z+3z^2+O(z^3)$. Keeping terms through $x^4$ givessoSince $d(\tan x)/dx={\sec }^{2}x$ and $\tan 0=0$, term-by-term integration supplies the constant of integration and givesThe expansions hold near zero; the nearest real poles at $\pm \pi /2$ limit their radius of convergence to $\pi /2$. The derivative relation also independently checks all three displayed coefficients of the tangent function.

The Maclaurin series give $1-\cos (x^2)=x^4/2+O(x^8)$ and $\tan x=x+O(x^3)$. Therefore $x^2{\tan }^{2}x=x^4+O(x^6)$, and cancellation of the common leading power givesThe argument applies from either side of zero: both leading terms involve the even power $x^4$. The expression need not be defined at zero itself for its limit to exist.

Combine the two singular terms before taking the limit:For $x$ near zero, within the real domain $x>-1$, the Maclaurin series is $\log (1+x)=x-x^2/2+O(x^3)$. HenceThe two individual terms have divergent limits; their leading singularities cancel. Evaluating them separately as an “infinity minus infinity” subtraction would not justify the finite result.

Subtract the Maclaurin series of $\log (1-x)$ from that of $\log (1+x)$. Their even powers cancel, givingThe full power series is $2{\sum }_{n\ge 0}{x}^{2n+1}/(2n+1)$, with radius of convergence one. For the real logarithm, the stated series applies on $|x|<1$.

For real $x$ sufficiently close to zero, $1+x>0$. Taking the logarithm and using its Maclaurin series givesContinuity of the exponential then givesThis is a two-sided real limit on the punctured neighbourhood where the base is positive.

The successive derivatives of cosine cycle through $\cos a,-\sin a,-\cos a,\sin a$. Its Taylor series about $a$ therefore separates into even and odd powers:The two brackets are the Maclaurin series of cosine and sine, givingBoth series converge absolutely for every real $x$, so regrouping the even and odd terms is valid everywhere on the real axis.

The hyperbolic cotangent isExpand the numerator and denominator using the exponential Maclaurin series:This is a Laurent series, rather than an ordinary Maclaurin series: $\coth x$ has a simple pole at zero. The printed request should be understood as the expansion obtained from the exponential series, including this unavoidable $x^{-1}$ term.

For $|x|<1$, the geometric series gives $(1+x{)}^{-1}={\sum }_{j=0}^{\infty }(-x{)}^j$. Integrating from zero to $x$ yields the Maclaurin seriesOn the real axis the series also converges at $x=1$ to $\ln 2$, but diverges at $x=-1$.

For fixed real $a$ and sufficiently large positive $x$, the base in the requested limit is positive. Apply the series with argument $-a^2/x^2$:Continuity of the exponential function therefore givesThis includes $a=0$, when the expression is identically one.

The Maclaurin series of the exponential function and sine giveBoth the numerator and denominator first become nonzero at order $x^4$. Dividing by that common power gives the limit

The fundamental theorem of calculus differentiates the variable upper endpoint:At zero, $I(0)=0$, $I^{\prime }(0)=2/\sqrt{\pi }$, $I^{\prime \prime }(0)=0$ and $I^{\prime \prime \prime }(0)=-4/\sqrt{\pi }$. The integrand is even, so its integral from zero is an odd function; all even coefficients in the Maclaurin series vanish. HenceThis is the local expansion of the error function.

Use the binomial series and exponential Maclaurin series:The Cauchy product isAdding $\log (1+x)=x-x^2/2+x^3/3-x^4/4+O(x^5)$ cancels the first three powers. Hence the first nonzero term is

The Maclaurin series $\cos x=1-x^2/2+O(x^4)$ gives $e^{\cos x}=e\exp (-x^2/2+O(x^4))$. The first two nonzero terms are therefore

To retain the finite term after cancellation, expand far enough before dividing. The Maclaurin series areThusand hence

The linear terms cancel: $\tan x=x+x^3/3+O(x^5)$, whereas $\tanh x=x-x^3/3+O(x^5)$. Also $\sinh x-x=x^3/6+O(x^5)$. Their leading nonzero terms therefore giveThe Maclaurin series remainder orders justify division after factoring out $x^3$.

The Maclaurin series give $\sin (3x)=3x+O(x^3)$ and $\sinh x=x+O(x^3)$. Dividing by $x$ in a punctured neighborhood of zero gives

The logarithm must be expanded about two, not about one. FactorApplying the supplied logarithm expansion to the bracket gives $\log 2+(x/2+x^2/4)-\frac{1}{2}(x/2{)}^2+O(x^3)$. Thus, through second order,In fact the cubic coefficient vanishes, since $\log (1+e^x)=x/2+\log [2\cosh (x/2)]$ and the second term is even; the Maclaurin series remainder can be sharpened to $O(x^4)$.