Fundamental limits as differences of reciprocals

The three fundamental limits (as they were taught to me) are

{\displaystyle \lim_{x\to 0} \frac{e^x-1}{x} = \lim_{x\to 0} \frac{\log(1+x)}{x} = \lim_{x\to 0} \frac{\sin x}{x} = 1}

(exponential, logarithmic, trigonometric). Usually they come with pictures like

standard-pic
exp(x) – 1 and x

which show that the graph of the numerator in each limit, say {y=f(x)}, is indeed close to the line {y=x}.

But there are so many different degrees of “close”: for example, {0.0103\approx 0.01} but the reciprocals of these numbers differ by almost {3}. As a stress-test of the approximation {f(x)\approx x}, let us consider the behavior of {\displaystyle \frac{1}{f(x)} - \frac{1}{x}} for each of the fundamental limits.

Exponential limit

Expressed as a difference: {\displaystyle \frac{1}{e^x - 1} - \frac{1}{x}}

expm1
1/(exp(x)-1) – 1/x

Always negative, which is obvious once we recall that the graph of {e^x-1} lies above its tangent line. The graph has central symmetry about {(0, -1/2)} which is not as obvious but is an easy algebra exercise. Since the asymptote on the right is {y=0}, the other asymptote is {y=-1}. This looks a lot like the logistic function {1/(1+\exp(-x))} … but it is not, because the logistic function approaches its asymptotes exponentially fast, while our function does so at the rate {1/x}.
For comparison, here is the logistic curve {\displaystyle y = \frac{1}{1+\exp(-x/3)} - 1} (in green) scaled to match the behavior at {0}.

expm1-compare
Too slow to be logistic

This could be potentially useful if one needs a logistic-type function that behaves like a rational function at infinity. Simply using a rational function for this purpose would not do: it cannot have two distinct horizontal asymptotes.

Logarithmic limit

{\displaystyle \frac{1}{\log(x+1)} - \frac{1}{x}}

log1p
1/log(1+x) – 1/x

This one is always positive, and has a vertical asymptote at {x=-1} in addition to the horizontal asymptote at {0}. At a glance it may look like shifted/scaled hyperbola {y=1/x}. Indeed, {y = (1+x/3)/(2+x)} is a decent approximation to it, shown in green below.

log1p-compare
Not really a hyperbola

Trigonometric limit

{\displaystyle \frac{1}{\sin x} - \frac{1}{x}}

sin
1/sin(x) – 1/x

Unlike in the previous cases, the difference of reciprocals vanished at {0} due to the approximation {\sin x \approx x} being of higher order: the error term is cubic rather than quadratic. The graph looks like the tangent function but it cannot be exactly that since the nearest vertical asymptotes are {x=\pm \pi} rather than {x=\pm \pi/2}. (Not to mention other reasons such as non-periodicity.) A stretched and rescaled tangent, namely {\displaystyle \frac{1}{3}\tan \frac{x}{2}}, sort of fits:

sin-compare
Like a tangent?

Bonus limit

{\tan x \approx x}, with the reciprocal difference being {\displaystyle \frac{1}{\tan x} - \frac{1}{x}}.

tan
1/tan(x) – 1/x

The limit {\displaystyle \lim_{x\to 0} \frac{\tan x}{x} = 1} adds nothing new compared to the previous one, since {\cos 0=1}. But the difference of reciprocals is another story. For one thing, the principal error term for it is twice as large as for the sine limit: {-x/3} versis {x/6}. Accordingly, the graph looks more like {\displaystyle -\frac{2}{3}\tan \frac{x}{2}}.

tan-compare
Not really like a tangent

Taylor series

All of the above differences have derivatives of all orders at the origin, which is not easy to prove with the standard calculus tools. Complex analysis makes the situation clear: the reciprocal of a holomorphic function with a zero at {z=0} can be expanded into a Laurent series, and subtracting {1/z} eliminates the principal part of that series, leaving a convergent power series, i.e., another holomorphic function. Let us take a look at these series:

{\displaystyle \frac{1}{e^x - 1} - \frac{1}{x} = - \frac{1}{2} + \frac{x}{12} - \frac{x^{3}}{720} + \frac{x^{5}}{30240} - \frac{x^{7}}{1209600} + \frac{x^{9}}{47900160} -\cdots }

Nice, the coefficients have alternating signs and all of them have numerator {1}… oh no, next term is {\displaystyle - \frac{691 x^{11}}{1307674368000}}. The signs do continue to alternate. Apart from the constant term, only odd powers of {x} appear, according to the central symmetry noted above. The coefficient of {x^{n-1}} in this series is {B_n/n!} where {B_n} is the {n}th Bernoulli number. These are the “modern” Bernoulli numbers, with {B_1 = -1/2} rather than {1/2}.

{\displaystyle \frac{1}{\log(x+1)} - \frac{1}{x} = \frac{1}{2} - \frac{x}{12} + \frac{x^{2}}{24} - \frac{19 x^{3}}{720} + \frac{3 x^{4}}{160} - \frac{863 x^{5}}{60480} + \cdots }

Also alternating but not omitting even powers, and not decaying nearly as fast as the coefficients of the previous series. (Of course: this function has a singularity at {-1} while the nearest singularities of the exponential thing are at {\pm 2\pi i}.)  These are Gregory coefficients, which according to Wikipedia “often appear in works of modern authors who do not recognize them”. I would not recognize them either without OEIS.

{\displaystyle \frac{1}{\sin x} - \frac{1}{x} = \frac{x}{6} + \frac{7 x^{3}}{360} + \frac{31 x^{5}}{15120} + \frac{127 x^{7}}{604800} + \frac{73 x^{9}}{3421440} + \cdots }

These are all positive for some reason.

{\displaystyle \frac{1}{\tan x} - \frac{1}{x} = - \frac{x}{3} - \frac{x^{3}}{45} - \frac{2 x^{5}}{945} - \frac{x^{7}}{4725} - \frac{2 x^{9}}{93555} - \cdots }

These are all negative for some reason.