Exponential and Logarithmic Limits

Plug x = 0 into \dfrac{e^x - 1}{x}. The numerator gives e^0 - 1 = 0. The denominator gives 0. You get \frac{0}{0} — undefined. The expression has no value at x = 0.

And yet, if you plug in x = 0.1, you get \frac{e^{0.1} - 1}{0.1} = \frac{0.10517\ldots}{0.1} = 1.0517\ldots Try x = 0.01: the result is 1.00502\ldots Try x = 0.001: the result is 1.000500\ldots The outputs are closing in on 1 from above.

Try from the other side. x = -0.1 gives \frac{e^{-0.1} - 1}{-0.1} = \frac{-0.09516\ldots}{-0.1} = 0.9516\ldots Then x = -0.01 gives 0.99502\ldots and x = -0.001 gives 0.999500\ldots The outputs close in on 1 from below.

From both sides, the value heads toward exactly 1. That is a limit — a number the expression approaches but never reaches at x = 0.

This single limit is the reason calculus can handle exponential functions at all. The derivative of e^x is e^x, and that fact rests entirely on this limit equalling 1. If it were anything other than 1, e^x would not be the remarkable function it is.

What about \frac{\ln(1 + x)}{x}? Plug in x = 0: the numerator is \ln 1 = 0, and the denominator is 0. Another \frac{0}{0}. Plug in x = 0.01: \frac{\ln(1.01)}{0.01} = \frac{0.00995\ldots}{0.01} = 0.995\ldots Heading toward 1 from below. The pattern repeats: four closely related expressions, all involving e^x or \ln x near zero, all producing \frac{0}{0} if you substitute directly, and all heading toward clean, specific values. These four limits form a family, and they depend on one another. Proving any one of them unlocks the rest.

Building the first limit from the definition of e

There is a number e \approx 2.71828\ldots that appears everywhere in mathematics. One of its many definitions is

This definition says: take \left(1 + \frac{1}{n}\right)^n for larger and larger n, and the values converge to a fixed number. That number is e.

The values are heading toward 2.71828\ldots — slowly, but steadily. This is e. The convergence is logarithmically slow: you need about 10 times more terms to get one extra correct decimal place. But the limit is definite — it converges to a fixed irrational number that sits between 2 and 3.

There is a concrete way to see why e appears. Suppose you deposit Rs 1 in a bank that offers 100% annual interest. If interest is compounded once a year, you end the year with Rs 2. Compounded twice a year (50% each half), you get (1 + 0.5)^2 = 2.25. Compounded monthly, (1 + 1/12)^{12} \approx 2.613. Compounded daily, (1 + 1/365)^{365} \approx 2.7146. The more often you compound, the closer you get to e \approx 2.71828. Continuous compounding — the limit as the number of compounding periods goes to infinity — gives you exactly e rupees.

A more flexible version of the same definition replaces 1/n with a continuous variable. Set t = 1/n, so as n \to \infty, t \to 0^+. Then

and by a symmetric argument (which you can verify by substituting t = -s and checking), the same limit holds as t \to 0^-. So

This is the continuous version of the definition, and it is the starting point for everything that follows.

The four standard limits

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

Proof. Let e^x - 1 = t, so e^x = 1 + t and x = \ln(1 + t). When x \to 0, the substitution gives t \to 0. The limit becomes

Rewrite the denominator. Since \ln(1+t) = \frac{1}{(1/t)}\ln(1+t)^{1/t}, you can write

As t \to 0, (1+t)^{1/t} \to e, so \ln\left((1+t)^{1/t}\right) \to \ln e = 1.

Why this matters: the derivative of e^x at x = 0 is exactly \lim_{h\to 0}\frac{e^{0+h} - e^0}{h} = \lim_{h\to 0}\frac{e^h - 1}{h}. This limit says that derivative is 1. That is the geometric statement that the tangent to y = e^x at the origin has slope exactly 1.

Limit 2: \displaystyle\lim_{x \to 0}\frac{\ln(1 + x)}{x} = 1

Proof. This is the reciprocal of what you just proved. Write

As x \to 0, (1+x)^{1/x} \to e, so

Why: this limit is equivalent to the first one. Setting t = e^x - 1 in Limit 1 gives exactly the expression in Limit 2. The two limits are two faces of the same fact — the exponential function and the logarithm are inverses, so their limiting behaviour near zero is linked.

Limit 3: \displaystyle\lim_{x \to 0}\frac{a^x - 1}{x} = \ln a (for a > 0, a \neq 1)

Proof. Write a^x = e^{x \ln a}, so a^x - 1 = e^{x\ln a} - 1. Let u = x \ln a. As x \to 0, u \to 0 as well. Then

By Limit 1, \frac{e^u - 1}{u} \to 1 as u \to 0, so

Why: this generalises Limit 1 from base e to any base a. When a = e, you get \ln e = 1, recovering Limit 1 as a special case. The key step is the identity a^x = e^{x\ln a}, which converts any exponential to base e.

Limit 4: \displaystyle\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e

This is the definition of e itself. But the result extends to a far more useful form. Replace 1/n with x/n:

Proof. Set m = n/x, so n = mx and \frac{x}{n} = \frac{1}{m}. As n \to \infty (with x fixed), m \to \infty. Then

Why: this is how compound interest works. If a bank offers 100% annual interest compounded n times per year, after one year you have \left(1 + \frac{1}{n}\right)^n rupees for every rupee deposited. As n \to \infty (continuous compounding), you get e rupees per rupee. Replace 100% with rate x, and you get e^x.

The substitution toolkit

Every exponential or logarithmic limit you meet in an exam is some disguised version of these four results. The key is recognising the disguise. Here are the standard substitutions.

Substitution 1: e^x - 1 = t. When the expression has e^x - 1 in it, set t = e^x - 1, so x = \ln(1 + t) and t \to 0 as x \to 0. This converts Limit 1 into Limit 2 and vice versa.

Substitution 2: a^x = e^{x \ln a}. Any exponential a^x can be rewritten in base e. This reduces Limit 3 to Limit 1.

Substitution 3: \ln(1+x) = t. When you see \ln(1+x), set t = \ln(1+x), so x = e^t - 1 and t \to 0 as x \to 0.

Substitution 4: reciprocal. For limits as x \to \infty involving (1 + 1/x)^x, set n = x and use the definition of e.

The key skill is not memorising these substitutions but recognising when to use each one. The test is simple: look at the indeterminate form. If you see e^{(\text{something})} - 1 divided by that same something, Limit 1 applies directly. If you see \ln(1 + \text{something}) divided by that something, Limit 2 applies. If the exponential has a base other than e, convert it to base e first. And if the whole expression has the shape of (1 + \text{small})^{\text{large}}, you are looking at the 1^\infty form.

Resolving the 1^\infty form

There is a class of limits that looks like this: as x \to a, you have a base approaching 1 and an exponent approaching infinity. The expression f(x)^{g(x)} takes the "form" 1^\infty. This is an indeterminate form — the result is not automatically 1.

The general technique: write

and evaluate \lim g(x) \ln f(x). If f(x) = 1 + h(x) where h(x) \to 0, then

By Limit 2, the last fraction tends to 1. So the whole expression tends to \lim g(x) \cdot h(x).

This gives a clean formula:

Why this works: the logarithm converts the power into a product, and then Limit 2 collapses the \frac{\ln(1 + h)}{h} factor to 1, leaving only g(x) \cdot h(x) to evaluate.

Formal definition

Worked examples

Common confusions

• "\lim\frac{e^x - 1}{x} equals e." A very common error. The limit is 1, not e. The number e lives in the base of the exponential, not in the value of the limit. The limit measures the slope of e^x at x = 0, and that slope happens to be 1.

• "Any expression of the form 1^\infty equals 1." False. 1^\infty is indeterminate — the answer depends on how fast the base approaches 1 relative to how fast the exponent grows. You just saw that \left(1 + \frac{3}{x}\right)^{2x} \to e^6, which is nowhere near 1.

• "The substitution e^x - 1 = t changes the limit point." The limit point changes from x \to 0 to t \to 0, yes — but the answer does not change. The substitution is a change of variable, not a change of problem. What matters is that t and x approach their limit points simultaneously.

• "Limit 3 works for a = 1." If a = 1, then a^x = 1 for all x, so a^x - 1 = 0 for all x. The expression \frac{0}{x} is 0 for all x \neq 0, so the limit is 0. And indeed \ln 1 = 0. So the formula still gives the right answer, but the proof divides by \ln a, which is 0 when a = 1. The result holds, but for a trivial reason.

• Confusing \ln(1+x)/x with \ln x / x. These are completely different limits. The first tends to 1 as x \to 0; the second is not even defined near x = 0 (since \ln x \to -\infty as x \to 0^+).

Going deeper

The series definition and a cleaner proof

There is another way to see Limit 1 that avoids the circular-looking substitution. The exponential function has a power series:

So

As x \to 0, every term after the 1 vanishes, leaving exactly 1. This proof is cleaner because it does not assume the continuous form of the definition of e — it uses the series definition instead. However, proving that the series converges to e^x requires its own careful argument, so the proof is not truly "simpler" — the difficulty has been moved, not removed.

The logarithmic series and Limit 2

Similarly, the Mercator series for the natural logarithm is

Dividing by x:

As x \to 0, this tends to 1. Madhava of Sangamagrama discovered a closely related series in the 14th century, well before the European development of calculus — an early instance of the deep connection between Indian mathematics and the theory of infinite series.

Combining the limits

Many competition problems chain several standard limits together. A useful identity:

This works because both limits exist individually (by Limit 3), so the limit of the difference is the difference of the limits.

Another useful combination: for limits of the form \frac{a^x - b^x}{c^x - d^x}, divide numerator and denominator by x:

Proving Limit 1 with the Sandwich Theorem

There is yet another proof of Limit 1, using the inequality 1 + x \leq e^x for all real x (which follows from the convexity of the exponential function — the tangent line at x = 0 lies below the curve). A related inequality, valid for x < 1, is e^x \leq \frac{1}{1 - x}. Together, these give:

For x > 0: from 1 + x \leq e^x, you get x \leq e^x - 1, so \frac{e^x - 1}{x} \geq 1.

From e^x \leq \frac{1}{1 - x} (for 0 < x < 1), you get e^x - 1 \leq \frac{x}{1-x}, so \frac{e^x - 1}{x} \leq \frac{1}{1 - x}.

Combining: 1 \leq \frac{e^x - 1}{x} \leq \frac{1}{1-x}. As x \to 0^+, both sides tend to 1, so by the Sandwich Theorem, \frac{e^x - 1}{x} \to 1. A similar argument works for x \to 0^-. This proof has the virtue of being entirely elementary — no series, no substitutions, just two inequalities and a squeeze.

The general 1^\infty form, carefully

The formula \lim f(x)^{g(x)} = e^{\lim g(x)\cdot h(x)} where f(x) = 1 + h(x) requires one subtlety: the limit \lim g(x) \cdot h(x) must exist (possibly as \pm\infty, in which case the answer is e^{\pm\infty} = \infty or 0). If g(x) \cdot h(x) oscillates without settling, the 1^\infty form has no limit.

The formula also assumes that f(x) > 0 near the limit point (so that f(x)^{g(x)} is defined). For exponential-type bases like 1 + k/x, this is automatic for large enough x.

A non-trivial example: evaluate \lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/x^2}. Here f(x) = \frac{\sin x}{x} \to 1 and g(x) = \frac{1}{x^2} \to \infty. Write f(x) = 1 + h(x) where h(x) = \frac{\sin x}{x} - 1. For small x, \sin x \approx x - \frac{x^3}{6}, so \frac{\sin x}{x} \approx 1 - \frac{x^2}{6}, giving h(x) \approx -\frac{x^2}{6}. Then g(x) \cdot h(x) \approx \frac{1}{x^2} \cdot \left(-\frac{x^2}{6}\right) = -\frac{1}{6}. So the limit is e^{-1/6}. The Taylor expansion of \sin x provided the key information about how fast the base deviates from 1.

Where this leads next

These four limits are the raw material for the next layer of calculus.

• Derivative — uses \lim \frac{e^h - 1}{h} = 1 to prove that the derivative of e^x is itself.
• Derivatives of Exponential and Logarithmic Functions — applies Limit 3 to find the derivative of a^x and Limit 2 to find the derivative of \ln x.
• Limits at Infinity — the behaviour of functions as x grows without bound, including exponential growth and decay.
• Sandwich Theorem — another technique for evaluating limits, especially useful for oscillating functions.
• Standard Limits — the complete catalogue of limits you need for JEE, including the trigonometric ones.