Standard Limits

Try computing \lim_{x \to 0} \frac{\sin x}{x}. Substitute x = 0 and you get \frac{0}{0}. Try factoring — but \sin x is not a polynomial, so there is nothing to factor. Try rationalising — there are no square roots. None of the algebraic tricks from the previous articles work. This limit needs a genuinely different argument.

And yet it comes up everywhere. Every derivative of a trigonometric function passes through it. Every Taylor series for \sin x rests on it. It appears in physics (wave interference), in engineering (signal processing), and in half the limit problems on a JEE paper. You need to know it cold — and more than that, you need to know why it equals 1, not just that it does.

This article proves five standard limits from first principles. Each one is an indeterminate form that resolves to a clean number through its own dedicated argument — geometric, algebraic, or a combination of both. Knowing these limits and their proofs is not optional — they are the foundation on which all of differential calculus is built.

Limit 1: \lim_{x \to 0} \frac{\sin x}{x} = 1

This is the most important standard limit. The proof is geometric — it uses areas and the squeeze theorem rather than any algebraic trick.

The geometric setup

Take a circle of radius 1 centred at the origin. Let x be a small positive angle (in radians) measured from the positive horizontal axis. Three quantities are connected to this angle:

1. \sin x — the height of the point on the circle above the horizontal axis. Geometrically, this is the vertical side of a right triangle inscribed in the circle.

2. x — the arc length from the horizontal axis to the point, which equals the angle in radians (since the radius is 1).

3. \tan x — the height of the point where the extended radius meets the vertical tangent line at the right end of the horizontal diameter.

The inequality

For 0 < x < \pi/2, compare the areas of three regions:

• Triangle OAB (inscribed in the arc) has area \frac{1}{2} \cdot 1 \cdot \sin x = \frac{\sin x}{2}.
• Circular sector OAB has area \frac{1}{2} \cdot 1^2 \cdot x = \frac{x}{2}.
• Triangle OAC (with the tangent line) has area \frac{1}{2} \cdot 1 \cdot \tan x = \frac{\tan x}{2}.

The triangle is inside the sector, and the sector is inside the larger triangle. So:

Multiply through by \frac{2}{\sin x} (positive, since \sin x > 0 for 0 < x < \pi/2):

Take reciprocals (flipping the inequalities):

The squeeze

Now let x \to 0^+. The left side \cos x \to \cos 0 = 1. The right side is already 1. By the squeeze theorem, the middle term is trapped:

For x \to 0^-, use the fact that \frac{\sin x}{x} = \frac{\sin(-x)}{-x} = \frac{\sin|x|}{|x|}, so the function is even and the left-sided limit equals the right-sided limit.

The requirement that x be in radians is essential. In degrees, \sin(1°) \approx 0.01745, so \frac{\sin 1°}{1} \approx 0.01745, which is nowhere near 1. The limit \frac{\sin x}{x} \to 1 is what makes radians the natural unit for angles in calculus. The radian is the angle unit where arc length equals angle times radius — and that geometric identity is precisely what the proof above used when it said "the arc length is x."

Here is a quick numerical check. Compute \frac{\sin x}{x} for a few small values (in radians):

The ratio is heading to 1 from below — never quite reaching it, but getting arbitrarily close. The squeeze theorem turns this numerical observation into a theorem.

Two useful consequences

From \frac{\sin x}{x} \to 1, two more limits follow quickly.

\lim_{x \to 0} \frac{\tan x}{x} = 1: Write \frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}. As x \to 0, the first factor \to 1 and the second \to 1/1 = 1. Product rule gives 1.

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}: Use the identity 1 - \cos x = 2\sin^2(x/2):

As x \to 0, x/2 \to 0, so \frac{\sin(x/2)}{x/2} \to 1. The limit is \frac{1}{2} \cdot 1^2 = \frac{1}{2}.

Limit 2: \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e

This limit defines the number e. It is a 1^\infty indeterminate form — the base 1 + \frac{1}{n} approaches 1, the exponent n approaches \infty, and the "rule" 1^\infty = 1 does not apply.

Why the limit exists

The sequence a_n = \left(1 + \frac{1}{n}\right)^n is increasing and bounded above, so it converges. Here is the argument.

Increasing: expand a_n using the binomial theorem:

Each factor \frac{n-j}{n} = 1 - \frac{j}{n} increases when n increases (with j fixed). So every term in the sum increases (or stays the same) when n increases, and the number of terms also increases. Hence a_n < a_{n+1}.

Bounded above: each factor \frac{n-j}{n} < 1, so each term is at most \frac{1}{k!}. Thus:

(using k! \geq 2^{k-1} for k \geq 1). So a_n < 3 for all n.

An increasing sequence bounded above must converge. The limit is the number e = 2.71828\ldots

The more general form — which appears constantly in JEE problems — replaces \frac{1}{n} with \frac{a}{n}:

This follows by substituting m = n/a: as n \to \infty, m \to \infty, and \left(1 + \frac{a}{n}\right)^n = \left[\left(1 + \frac{1}{m}\right)^m\right]^a \to e^a.

Limit 3: \lim_{x \to 0} \frac{e^x - 1}{x} = 1

This limit says: for small x, e^x \approx 1 + x. It is the first-order approximation of the exponential function, and it is the reason e^x is its own derivative.

Proof from the definition of e

Set e^x - 1 = t, so e^x = 1 + t and x = \ln(1 + t). As x \to 0, t \to 0. Then:

The reciprocal is \frac{\ln(1+t)}{t}. If this limit equals 1 (proved below as Limit 5), then \frac{t}{\ln(1+t)} \to 1 as well. So the two limits — \frac{e^x - 1}{x} and \frac{\ln(1+x)}{x} — are mutual reciprocals, and both equal 1.

An independent argument

Alternatively, use the definition e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n to establish e^x - 1 \approx x for small x.

For a positive integer n, set x = 1/n. Then e^x \approx \left(1 + \frac{1}{n}\right) for large n (the n-th root of e is approximately 1 + 1/n). So e^x - 1 \approx \frac{1}{n} = x, giving \frac{e^x - 1}{x} \approx 1. This is not a rigorous proof, but it shows why the limit should be 1.

This limit has a beautiful geometric reading: the curve y = e^x passes through the point (0, 1) with slope exactly 1. Among all exponential curves y = a^x, the base e is the unique one whose slope at x = 0 is 1. That is the defining property of e.

A quick numerical check confirms the limit:

The ratio approaches 1 from above — the higher-order terms in the Taylor expansion e^x = 1 + x + \frac{x^2}{2} + \cdots contribute a small positive excess that vanishes as x \to 0.

Limit 4: \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a, for a > 0

This is the generalisation of Limit 3 to any base a, not just e.

Proof

Write a^x = e^{x \ln a}. Let u = x \ln a. As x \to 0, u \to 0. Then:

As x \to 0, u \to 0, so \frac{e^u - 1}{u} \to 1 by Limit 3. The product rule gives:

Check: when a = e, \ln a = 1, recovering Limit 3. When a = 2, the limit is \ln 2 \approx 0.693. When a = 10, the limit is \ln 10 \approx 2.303.

This limit tells you the rate at which exponential functions start to grow. At x = 0, the function a^x passes through the point (0, 1) with slope \ln a. The "natural" base e is the unique base for which this initial slope is exactly 1 — that is what makes e natural.

Limit 5: \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1

This is the logarithmic counterpart of Limit 3: for small x, \ln(1 + x) \approx x.

Proof

Write \frac{\ln(1+x)}{x} = \frac{1}{x} \cdot \ln(1+x) = \ln(1+x)^{1/x}.

Let n = 1/x. As x \to 0^+, n \to \infty, and:

By Limit 2, \left(1 + \frac{1}{n}\right)^n \to e, and since \ln is continuous:

For x \to 0^-, a similar substitution works: let x = -1/(m+1) where m \to \infty, and verify that the limit is again 1.

This says \ln(1 + x) \approx x for small x. It is the inverse of the approximation e^x \approx 1 + x. In fact, these two approximations are the same statement from opposite directions: if e^x \approx 1 + x, then applying \ln to both sides gives x \approx \ln(1 + x).

A numerical check:

The ratio approaches 1 from below — the higher-order terms in \ln(1+x) = x - \frac{x^2}{2} + \cdots subtract a small amount that vanishes as x \to 0.

How the five limits connect

The five standard limits are not isolated facts — they form a tightly linked web. Limit 3 (\frac{e^x - 1}{x}) and Limit 5 (\frac{\ln(1+x)}{x}) are reciprocals of each other. Limit 4 (\frac{a^x - 1}{x}) reduces to Limit 3 by writing a^x = e^{x \ln a}. Limit 5 reduces to Limit 2 (the definition of e) by exponentiating. And Limit 1 (\frac{\sin x}{x}) stands on its own geometric foundation, independent of the others.

Once you know any two of Limits 2, 3, and 5, the third follows. The choice of which to treat as "fundamental" is a matter of taste. This article derives Limit 2 first (from the binomial theorem), uses it to prove Limit 5, and then derives Limit 3 from Limit 5.

Putting it all together

Common confusions

• "\frac{\sin x}{x} = 1 always." Only as x \to 0. For any fixed x \neq 0, \frac{\sin x}{x} is a definite number less than 1 (for 0 < x < \pi) or could be negative. The standard limit is a statement about what happens in the limit, not an identity.

• "\frac{\sin x}{x} \to 1 works in degrees." It does not. If x is measured in degrees, the limit is \frac{\pi}{180} \approx 0.01745, not 1. The geometric proof uses the fact that a radian is the angle whose arc length equals the radius — that connection breaks if you use degrees. Calculus always uses radians.

• "\lim \frac{e^x - 1}{x} = 1 means e^x - 1 = x." Close, but no. e^x - 1 \approx x for x near 0 is a good approximation, not an equality. The exact expansion is e^x - 1 = x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots, and all those higher-order terms are negligible compared to x when x is small — which is exactly why the limit works out to 1.

• "I can use these limits with the argument in degrees." All five standard limits assume the natural conventions: angles in radians, logarithms natural (base e). If you use degrees or \log_{10}, conversion factors appear.

• "\lim_{x \to 0} \frac{\sin 5x}{x} = 1." Wrong — the argument of \sin and the denominator must match. Here \lim_{x \to 0} \frac{\sin 5x}{x} = 5. The general pattern is \lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b}.

Going deeper

More derived limits

The five standard limits generate many others through algebraic manipulation. Here are the most frequently used:

\lim_{x \to 0} \frac{\tan x}{x} = 1. Proved above by writing \tan x = \frac{\sin x}{\cos x}.

\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}. Proved above using the half-angle identity.

\lim_{x \to 0} \frac{1 - \cos x}{x} = 0. Write \frac{1 - \cos x}{x} = \frac{1-\cos x}{x^2} \cdot x. The first factor \to \frac{1}{2} and the second \to 0, so the product is 0.

\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1. Let x = \sin \theta, so \theta = \sin^{-1} x. As x \to 0, \theta \to 0. Then \frac{\sin^{-1} x}{x} = \frac{\theta}{\sin \theta} = \frac{1}{\frac{\sin \theta}{\theta}} \to \frac{1}{1} = 1.

\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1. Similarly, let x = \tan \theta, \theta = \tan^{-1} x. Then \frac{\tan^{-1} x}{x} = \frac{\theta}{\tan \theta} \to 1.

\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n (for any real n). This is because (1+x)^n \approx 1 + nx for small x (the binomial approximation). More precisely:

The first factor \to 1 by Limit 3 (as n\ln(1+x) \to 0) and the second \to n \cdot 1 = n by Limit 5.

The Taylor series connection

Each standard limit is really the statement that a certain function, divided by x, has its first-order Taylor coefficient equal to 1 (or \ln a). In the language of Taylor series:

Once you know Taylor series (covered in a later article), all five standard limits become immediate. But the Taylor series themselves rest on derivatives, and derivatives rest on these limits — so there is a pleasant circularity here. The geometric proof of \frac{\sin x}{x} \to 1 breaks the circle by proving the limit independently of derivatives.

A table for reference

Where this leads next

These five limits are the raw materials for most of differentiation. With them in hand, you can derive:

• Derivative — the definition and first-principles computation of derivatives. The limit \frac{\sin x}{x} \to 1 is what makes \frac{d}{dx}\sin x = \cos x true.
• Trigonometric Derivatives — how to differentiate \sin x, \cos x, \tan x, and the rest, using the standard limits as building blocks.
• L'Hôpital's Rule — the general tool for \frac{0}{0} and \frac{\infty}{\infty} forms, which subsumes all the algebraic tricks and the standard limits into one powerful theorem.
• Taylor Series — the infinite polynomial approximation of a function, which explains why \sin x \approx x and e^x \approx 1 + x near the origin.
• Indeterminate Forms — the full catalogue of cases where substitution fails, and the strategies for resolving each one.