Limits Using Expansion

Try to compute this limit directly:

Plug in x = 0. The numerator gives 1 - 1 - 0 = 0. The denominator gives 0. You get 0/0 -- nothing you can cancel, nothing you can factor. The standard limit tricks from earlier articles do not apply here because there is no single standard form hiding inside.

But there is a way through. What if you knew that, for small x,

Then the numerator becomes

Divide by x^2:

Now let x \to 0. Every term with an x in it vanishes, leaving \dfrac{1}{2}.

That is the entire method. Replace each function by its series, do the algebra, and the limit appears.

Compare this with the alternative approaches. You could try L'Hopital's rule, but you would need to differentiate twice (the first round gives \frac{e^x - 1}{2x}, still 0/0). You could try algebraic manipulation, but there is nothing to factor. The expansion cuts through all of that: one substitution, one simplification, done.

The rest of this article is about learning which series to use, why they work, and how to handle the algebra cleanly.

Where the expansions come from

The series e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots is not a guess. It is a theorem: under mild conditions, a function can be written as an infinite sum of powers of x, with each coefficient determined by the function's derivatives at x = 0. This infinite sum is called the Maclaurin series of the function (or more generally, a Taylor series if the expansion is centred at some other point).

The general formula is

For limits, you do not need to derive this formula from scratch every time. You only need the standard expansions -- a short list of series for the functions that appear most often.

The standard expansions

Here are the six expansions you will use in almost every problem. All are valid for x near 0.

The last one is the binomial series for any real exponent n -- the same binomial theorem you know for positive integers, extended to fractions, negatives, and irrationals.

Notice the patterns. The sine series has only odd powers. The cosine series has only even powers. The exponential series has all powers with all-positive signs and factorials in the denominator. The logarithm series has all powers with alternating signs and plain integers in the denominator. Knowing these patterns by sight is half the skill.

The technique, step by step

Every expansion-based limit follows the same four-step process:

1. Identify the small quantity. In \lim_{x \to 0}, the small quantity is x. In \lim_{x \to a}, substitute h = x - a and expand in powers of h.

2. Expand each function using the standard series. Keep enough terms so that the leading term in the numerator does not cancel completely — usually two or three terms beyond the first surviving one.

3. Simplify the numerator and denominator. Cancel common factors, combine like terms.

4. Take the limit. After cancellation, the expression will be a polynomial plus higher-order terms. The limit is the constant term (or the coefficient of the lowest surviving power).

The key judgement call is how many terms to keep. There is a reliable rule: look at the denominator. If the denominator is x^2, you need the numerator expanded to at least the x^2 term. If the denominator is x^3, expand to the x^3 term. Keep one extra term as a safety margin.

A quick example to see the process in action. Evaluate \lim_{x \to 0} \frac{1 - \cos x}{x^2}. The denominator is x^2, so expand \cos x to the x^2 term: \cos x = 1 - x^2/2 + \cdots. Then 1 - \cos x = x^2/2 + \cdots, and dividing by x^2 gives 1/2 + \cdots \to 1/2. One line of expansion, one line of division, done.

Comparing orders

When two functions are subtracted and both are close to the same value near x = 0, the leading terms cancel and the limit depends on the first term that differs. This is called comparing orders.

Take \sin x and x. For small x:

So \sin x - x = -\frac{x^3}{6} + \cdots. The two functions agree up to order x. They first disagree at order x^3. That disagreement is what determines the limit of \frac{\sin x - x}{x^3}.

This is the power of the method: by comparing the order at which two functions first differ, you immediately know the answer.

Reading the definition. The series does the work that factoring cannot. In a 0/0 limit, the numerator and denominator both vanish at x = 0, which means their series both start with a positive power of x, not a constant. Dividing cancels the shared power, and the constant term of what remains is the limit.

A first worked example

A second example, with logarithm and cosine

Common confusions

• "I need to memorise the full infinite series." You do not. For limits, you almost never need more than the first three or four terms. The rest are absorbed into the "+ \cdots" and vanish when you take the limit.

• "The expansion method only works when x \to 0." It works for any limit, but you may need to substitute first. For \lim_{x \to 1}, set h = x - 1 so that h \to 0, then expand in powers of h. For \lim_{x \to \infty}, set t = 1/x so that t \to 0.

• "If the numerator cancels to zero, the limit is zero." Only if the numerator cancels to a higher power of x than the denominator. If both cancel to the same power, the limit is a finite nonzero number (like Example 2). If the numerator cancels to a lower power, the limit is infinite.

• "I should expand everything to the same number of terms." Not quite. You should expand everything to the same order of x. The sine series has no x^2 term, so expanding \sin x to three terms gives x - x^3/6 + x^5/120, which is already at order x^5. Meanwhile, expanding e^x to three terms gives 1 + x + x^2/2, only at order x^2. Match the power, not the number of terms.

• "Why not just use L'Hopital's rule?" You can, and for simple limits L'Hopital is faster. But for limits where the 0/0 survives after one or two differentiations, the expansion method is cleaner. It also reveals why a particular limit has the value it does -- which power of x survives -- while L'Hopital just produces a number.

Going deeper

Why the series works: Taylor's theorem with remainder

The Maclaurin series is not magic. It is guaranteed by a theorem.

Taylor's theorem says: if f has n + 1 continuous derivatives near x = 0, then

where the remainder R_n(x) satisfies R_n(x)/x^n \to 0 as x \to 0. The remainder is the error from truncating the series, and it is small enough that it does not affect the limit.

This is why the method works: when you write e^x = 1 + x + x^2/2 + \cdots and throw away the higher terms, the error you are making vanishes faster than the terms you are keeping. So the limit of the truncated expression is the same as the limit of the full expression.

The remainder can be written explicitly as

for some c between 0 and x. This is the Lagrange form of the remainder, and it gives you a bound on the error. For our purposes, all that matters is that R_n(x) = O(x^{n+1}) -- it is of order x^{n+1}, meaning it vanishes at least as fast as x^{n+1}.

Deriving the standard expansions

Where do the specific series come from? Take e^x as the example. All its derivatives are e^x, which is 1 at x = 0. So

and the series is \sum \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots.

For \sin x: the derivatives cycle through \sin, \cos, -\sin, -\cos, \sin, \ldots At x = 0 these are 0, 1, 0, -1, 0, 1, \ldots So only the odd-powered terms survive, with alternating signs.

For \ln(1 + x): the first derivative is (1+x)^{-1} = 1 at x = 0. The second derivative is -(1+x)^{-2} = -1 at x = 0. The third is 2(1+x)^{-3} = 2 at x = 0. In general, f^{(n)}(0) = (-1)^{n-1}(n-1)!, giving the series x - x^2/2 + x^3/3 - \cdots.

The substitution trick for non-zero limits

When the limit is at x = a instead of x = 0, substitute h = x - a so that h \to 0. Then rewrite everything in terms of h and expand.

For instance, \displaystyle\lim_{x \to \pi/2} \frac{\cos x}{x - \pi/2}. Set h = x - \pi/2, so x = \pi/2 + h and h \to 0.

So the limit becomes

The substitution reduces every limit-at-a-point to a limit at zero, where the standard expansions apply directly.

Expanding compositions

When a limit involves a function of a function -- say \sin(\tan x) or e^{\sin x} -- you expand from the inside out.

Take \lim_{x \to 0} \frac{e^{\sin x} - e^x}{x^3}. First, expand \sin x:

So \sin x = x + u where u = -x^3/6 + \cdots. Now expand e^{\sin x} = e^{x + u} = e^x \cdot e^u. For small u, e^u \approx 1 + u + u^2/2 + \cdots = 1 + u + \cdots (since u^2 is order x^6, too small to matter here).

Divide by x^3: \frac{e^{\sin x} - e^x}{x^3} \to -\frac{e^0}{6} = -\frac{1}{6}.

The trick is to keep track of which terms matter. Since the denominator is x^3, you only need the numerator to order x^3. The inside expansion \sin x = x - x^3/6 + \cdots gives the correction at exactly the right order.

When the expansion approach is the best tool

The expansion method is especially powerful when:

• Multiple cancellations happen. If the numerator is a difference of two functions that agree up to order x^k, you need k+1 terms of each expansion. No other technique reveals this structure as cleanly.

• The limit involves compositions. For \lim_{x \to 0} \frac{\sin(\tan x) - \tan(\sin x)}{x^7}, a famous competition problem, expanding step by step is the only practical approach. (The answer, for the curious, is -1/30.)

• You need to determine the exact order of vanishing. Knowing that \sin x - x is of order x^3 while \tan x - x is of order x^3 but with coefficient +1/3 instead of -1/6 is information that L'Hopital does not naturally provide.

A quick-reference table of leading behaviours

These approximations, all valid for small x, are worth knowing by heart. They tell you the order at which each expression vanishes or differs from a simpler one.

When you see one of these expressions in a limit, you can immediately read off the order and the coefficient, which is often all you need.

Where this leads next

The expansion method is one of three major tools for indeterminate limits. The others complement it:

• L'Hopital's Rule -- the differentiation shortcut for 0/0 and \infty/\infty. Faster for simple limits, but less revealing than expansions for limits with deep cancellation.
• Special Limit Forms -- limits of the form 1^\infty, 0^0, and \infty^0, which need their own techniques.
• Indeterminate Forms -- the full catalogue of 0/0, \infty/\infty, 0 \cdot \infty, and the others, with strategies for each.
• Binomial Theorem for Rational Index -- the expansion (1+x)^n for any real n, which is the engine behind many of these series.
• Derivative -- the Maclaurin series encodes all the derivatives of a function at a single point. The two ideas are deeply connected.