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Special Limits



Why Memorize Special Limits?
The Fundamental Trigonometric Limit
Related Trigonometric Limits
Deriving (1 − cos x)/x = 0
The Natural Exponential Limit
Related Exponential Limits
The Definition of e
Limits Involving Logarithms
Growth Rate Comparisons
Using Special Limits
Summary: Master Reference of Special Limits
Special Limits FAQ



Limits Worth Memorizing


Certain limits appear so frequently in calculus that recognizing them on sight saves considerable effort. Each of these limits yields an indeterminate form under direct substitution—typically 0/00/0—yet each has a definite, well-established value.

These special limits are not mere curiosities. The limit sin⁡xx→1\dfrac{\sin x}{x} \to 1 underlies the derivatives of all trigonometric functions. The limit ex−1x→1\dfrac{e^x - 1}{x} \to 1 defines what makes the exponential function special. The limit (1+1/x)x→e(1 + 1/x)^x \to e provides one definition of ee itself.

Knowing these limits transforms difficult calculations into straightforward applications. When a complicated expression can be massaged into a form matching one of these patterns, the work is essentially done.

Key Terms

Limit— the object whose value these results establish
Indeterminate Form— each special limit arises from an indeterminate form

See All Calculus Definitions →


Why Memorize Special Limits?


    Special limits resist direct computation. Substitution yields indeterminate forms, and algebraic manipulation alone cannot resolve them without circular reasoning or advanced tools.

    Their values come from geometric arguments, series expansions, or the definitions of the functions involved. Once established, these limits become permanent tools in the calculus toolkit.

    Memorizing them provides immediate payoff:

  • Faster problem-solving when these forms appear
  • Building blocks for evaluating more complex limits
  • Foundation for understanding derivatives of transcendental functions
  • Recognition of patterns that guide algebraic manipulation


The Fundamental Trigonometric Limit


Sine Limit at Zero
lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
Learn more about this formula: Sine Limit at Zero →


This limit requires xx measured in radians. Direct substitution gives 0/00/0, revealing nothing.

The standard proof uses the unit circle. For small positive xx, three quantities satisfy:

The Fundamental Trigonometric Limit
sin⁡x<x<tan⁡x\sin x < x < \tan x

Dividing by sin⁡x\sin x:

The Fundamental Trigonometric Limit
1<xsin⁡x<1cos⁡x1 < \frac{x}{\sin x} < \frac{1}{\cos x}

Taking reciprocals and applying the Squeeze Theorem:

The Fundamental Trigonometric Limit
cos⁡x<sin⁡xx<1\cos x < \frac{\sin x}{x} < 1

As x→0x \to 0, cos⁡x→1\cos x \to 1, so sin⁡xx\dfrac{\sin x}{x} is squeezed to 11.

This limit determines the derivative of sin⁡x\sin x: the fact that (sin⁡x)′=cos⁡x(\sin x)' = \cos x depends entirely on this result.
sin x / x lies between cos x and 1 near 0 and has a hole at (0, 1): its limit is 1xy−3−2−1123−11y = 1cos xsin x / xcos x < sin x / x < 1 near 0hole at (0, 1)Squeezed between cos x and 1, sin x / x must approach 1
sin x / x between cos x and 1

The section’s squeeze, drawn. Near 0 (in radians) the curve sin x / x lies above cos x and below the dashed line y = 1; at x = 0 itself it has no value, only a hole at (0, 1). Both bounds tend to 1 as x → 0, so the trapped curve tends to 1 as well. That is why lim sin x / x = 1 even though substituting 0 gives 0/0.

In degrees the same quotient tends to π/180, which is why calculus uses radians.

Related Trigonometric Limits


Several limits follow from the fundamental trigonometric limit.

Cosine Limit at Zero
lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0
Learn more about this formula: Cosine Limit at Zero →

Cosine Quadratic Limit at Zero
lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}
Learn more about this formula: Cosine Quadratic Limit at Zero →

Tangent Limit at Zero
lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1
Learn more about this formula: Tangent Limit at Zero →


Related Trigonometric Limits
lim⁡x→0xsin⁡x=1\lim_{x \to 0} \frac{x}{\sin x} = 1

The last follows by taking the reciprocal: if sin⁡xx→1\dfrac{\sin x}{x} \to 1, then xsin⁡x→1\dfrac{x}{\sin x} \to 1 as well.
Limit (x → 0) Value Note
sin x / x 1 the fundamental trig limit; x in radians
(1 − cos x) / x 0 derived via conjugate trick (see obj4 below)
(1 − cos x) / x² 1/2 sharper version with higher-order denominator
tan x / x 1 factor as (sin x / x) · (1 / cos x), each → 1
x / sin x 1 reciprocal of the fundamental trig limit

Deriving (1 − cos x)/x = 0


Start with the expression and multiply by the conjugate:

Deriving (1 − cos x)/x = 0
1−cos⁡xx⋅1+cos⁡x1+cos⁡x=1−cos⁡2xx(1+cos⁡x)=sin⁡2xx(1+cos⁡x)\frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x(1 + \cos x)} = \frac{\sin^2 x}{x(1 + \cos x)}

Rewrite as a product:

Deriving (1 − cos x)/x = 0
=sin⁡xx⋅sin⁡x1+cos⁡x= \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x}

As x→0x \to 0:

Deriving (1 − cos x)/x = 0
sin⁡xx→1sin⁡x1+cos⁡x→02=0\frac{\sin x}{x} \to 1 \qquad \frac{\sin x}{1 + \cos x} \to \frac{0}{2} = 0

Therefore:

Deriving (1 − cos x)/x = 0
lim⁡x→01−cos⁡xx=1⋅0=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 1 \cdot 0 = 0

The Natural Exponential Limit


Exponential Limit at Zero
lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1
Learn more about this formula: Exponential Limit at Zero →


Direct substitution gives 0/00/0. This limit defines the derivative of exe^x at x=0x = 0:

The Natural Exponential Limit
ddxex∣x=0=lim⁡h→0e0+h−e0h=lim⁡h→0eh−1h=1\frac{d}{dx} e^x \bigg|_{x=0} = \lim_{h \to 0} \frac{e^{0+h} - e^0}{h} = \lim_{h \to 0} \frac{e^h - 1}{h} = 1

THEOREM
The exponential function is the unique function satisfying f′(x)=f(x)f'(x) = f(x) with f(0)=1f(0) = 1.

This limit is the cornerstone of that property.

An equivalent form:

The Natural Exponential Limit
lim⁡h→0eh−1h=1\lim_{h \to 0} \frac{e^h - 1}{h} = 1

Related Exponential Limits


For any base a>0a > 0:

Related Exponential Limits
lim⁡x→0ax−1x=ln⁡a\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a

When a=ea = e, this reduces to the natural exponential limit since ln⁡e=1\ln e = 1.

For the natural logarithm:

Logarithm Taylor Limit
lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1
Learn more about this formula: Logarithm Taylor Limit →


This can be seen by substituting u=ln⁡(1+x)u = \ln(1 + x), so eu=1+xe^u = 1 + x and x=eu−1x = e^u - 1. As x→0x \to 0, u→0u \to 0:

Related Exponential Limits
lim⁡x→0ln⁡(1+x)x=lim⁡u→0ueu−1=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{u \to 0} \frac{u}{e^u - 1} = 1

The Definition of e


The number ee emerges from limits:

Definition of e as a Limit
lim⁡x→∞(1+1x)x=e\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e
Learn more about this formula: Definition of e as a Limit →


Equivalently, using discrete notation:

The Definition of e
lim⁡n→∞(1+1n)n=e\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e

Another form with x→0x \to 0:

The Definition of e
lim⁡x→0(1+x)1/x=e\lim_{x \to 0} (1 + x)^{1/x} = e

The value is e≈2.71828e \approx 2.71828. This limit arises naturally in compound interest: if interest is compounded nn times per year at annual rate 100%100\%, the growth factor over one year is (1+1/n)n(1 + 1/n)^n, which approaches ee as n→∞n \to \infty.
(1 + 1/n)^n for n = 1 to 12: 2, 2.25, 2.37, ... rising toward e = 2.718ny123456789101112132e ≈ 2.718n = 1: 2n = 2: 2.25n = 12: 2.613(1 + 1/n)ⁿ climbs toward e but never reaches it
Compounding toward e

The terms (1 + 1/n)ⁿ for n = 1 to 12: 2, 2.25, 2.370, 2.441, …, 2.613. Each one is the growth factor of a year at 100% interest compounded n times. More frequent compounding always helps, so the dots keep rising, but by less and less, and they stay below the dashed level e ≈ 2.718, which they approach as n grows without bound.

The approach is slow: at n = 1000 the value is still 2.7169, short of e by about 0.0014.

Limits Involving Logarithms


Logarithms grow slowly—slower than any positive power of xx.

x ln x Limit at Zero
lim⁡x→0+xln⁡x=0\lim_{x \to 0^+} x \ln x = 0
Learn more about this formula: x ln x Limit at Zero →


This is a 0⋅(−∞)0 \cdot (-\infty) form. As x→0+x \to 0^+, xx vanishes while ln⁡x→−∞\ln x \to -\infty. The factor xx wins: the product approaches 00.

Limits Involving Logarithms
lim⁡x→∞ln⁡xx=0\lim_{x \to \infty} \frac{\ln x}{x} = 0

As x→∞x \to \infty, both numerator and denominator grow, but xx grows faster than ln⁡x\ln x. The ratio vanishes.

More generally, for any n>0n > 0:

Limits Involving Logarithms
lim⁡x→∞ln⁡xxn=0lim⁡x→0+xnln⁡x=0\lim_{x \to \infty} \frac{\ln x}{x^n} = 0 \qquad \lim_{x \to 0^+} x^n \ln x = 0

Logarithms lose to any positive power.

Growth Rate Comparisons


As x→∞x \to \infty, functions grow at different rates. The hierarchy:

Growth Rate Comparisons
logarithmic≪polynomial≪exponential\text{logarithmic} \ll \text{polynomial} \ll \text{exponential}

Specifically:

Polynomial Beats Logarithm
lim⁡x→∞ln⁡xxn=0,n>0\lim_{x \to \infty} \frac{\ln x}{x^n} = 0, \quad n > 0
Learn more about this formula: Polynomial Beats Logarithm →

Exponential Beats Polynomial
lim⁡x→∞xnex=0\lim_{x \to \infty} \frac{x^n}{e^x} = 0
Learn more about this formula: Exponential Beats Polynomial →


Growth Rate Comparisons
lim⁡x→∞xnax=0for any a>1 and any n\lim_{x \to \infty} \frac{x^n}{a^x} = 0 \quad \text{for any } a > 1 \text{ and any } n

Exponentials dominate polynomials, which dominate logarithms. These comparisons determine which terms control behavior in limits at infinity.
Pair Slower grower Faster grower Limit (x → ∞)
Logarithmic vs polynomial ln x xn, any n > 0 ln x / xn → 0
Polynomial vs natural exponential xn, any n ex xn / ex → 0
Polynomial vs general exponential xn, any n ax, any a > 1 xn / ax → 0

Using Special Limits


Rewrite expressions to match known forms.

Example 1


Using Special Limits
lim⁡x→0sin⁡5x3x\lim_{x \to 0} \frac{\sin 5x}{3x}

Rewrite to expose the standard form:

Using Special Limits
=53⋅lim⁡x→0sin⁡5x5x=53⋅1=53= \frac{5}{3} \cdot \lim_{x \to 0} \frac{\sin 5x}{5x} = \frac{5}{3} \cdot 1 = \frac{5}{3}

Example 2


Using Special Limits
lim⁡x→0e4x−1x\lim_{x \to 0} \frac{e^{4x} - 1}{x}

Factor out the coefficient:

Using Special Limits
=4⋅lim⁡x→0e4x−14x=4⋅1=4= 4 \cdot \lim_{x \to 0} \frac{e^{4x} - 1}{4x} = 4 \cdot 1 = 4

Example 3


Using Special Limits
lim⁡x→0tan⁡xx=lim⁡x→0sin⁡xx⋅1cos⁡x=1⋅1=1\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1

Recognizing patterns and factoring to match special limits streamlines computation.

Summary: Master Reference of Special Limits


Every limit on this page appears repeatedly throughout calculus, and pattern-matching against them is the first move for any indeterminate-form limit. The table below collects them in one master reference, grouped by family — trigonometric, exponential, the definition of e, boundary-behavior logarithmic limits, and growth-rate comparisons as x → ∞. The first useful question to ask about an unfamiliar limit is whether it can be massaged into one of these forms; very often, the answer is yes.
Family Limit Value
Trigonometric limx → 0 sin x / x 1
Trigonometric limx → 0 (1 − cos x) / x 0
Trigonometric limx → 0 (1 − cos x) / x² 1/2
Trigonometric limx → 0 tan x / x 1
Trigonometric limx → 0 x / sin x 1
Exponential limx → 0 (ex − 1) / x 1
Exponential limx → 0 (ax − 1) / x  (a > 0) ln a
Exponential limx → 0 ln(1 + x) / x 1
Definition of e limx → ∞ (1 + 1/x)x = limx → 0 (1 + x)1/x e
Logarithmic (boundary) limx → 0⁺ x ln x 0
Logarithmic (boundary) limx → 0⁺ xn ln x  (n > 0) 0
Growth-rate (x → ∞) limx → ∞ ln x / xn  (n > 0) 0
Growth-rate (x → ∞) limx → ∞ xn / ex  (any n) 0
Growth-rate (x → ∞) limx → ∞ xn / ax  (any n, a > 1) 0

Special Limits FAQ

What is the limit of sin(x)/x as x approaches 0?

+
The limit equals 1, with x measured in radians. Direct substitution gives 0/0, so the standard proof uses the unit circle inequality sin x < x < tan x, which rearranges to cos x < sin(x)/x < 1. Since cos x approaches 1, the Squeeze Theorem forces the middle term to 1. This result is what makes the derivative of sin x equal cos x.Read more →

What is the limit of (e^x − 1)/x as x approaches 0?

+
The limit equals 1. Direct substitution gives 0/0, but the expression is exactly the difference quotient for e^x at x = 0, so the limit is the derivative of e^x there. Its value of 1 is what makes e the unique base whose exponential function is its own derivative, with f(0) = 1.Read more →

What is the limit of ln(1+x)/x as x approaches 0?

+
The limit equals 1. Substituting u = ln(1 + x) gives x = e^u − 1, and as x approaches 0 so does u, turning the expression into u/(e^u − 1) — the reciprocal of the natural exponential limit, which is also 1. The two results are the same fact viewed from opposite sides.Read more →

What is the limit of (a^x − 1)/x as x approaches 0?

+
For any base a > 0, the limit equals ln a. This generalizes the natural exponential limit: setting a = e gives ln e = 1, recovering the familiar result. The appearance of ln a here is why the derivative of a^x is a^x·ln a rather than simply a^x, and why base e is the convenient choice.Read more →