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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



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.



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:



The Fundamental Trigonometric Limit


lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1


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:

sin⁡x<x<tan⁡x\sin x < x < \tan x


Dividing by sin⁡x\sin x:

1<xsin⁡x<1cos⁡x1 < \frac{x}{\sin x} < \frac{1}{\cos x}


Taking reciprocals and applying the Squeeze Theorem:

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.

Related Trigonometric Limits


Several limits follow from the fundamental trigonometric limit.

lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0


lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}


lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1


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.

Deriving (1 − cos x)/x = 0


Start with the expression and multiply by the conjugate:

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:

=sin⁡xx⋅sin⁡x1+cos⁡x= \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x}


As x→0x \to 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:

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


lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1


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

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


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:

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


Related Exponential Limits


For any base a>0a > 0:

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:

lim⁡x→0ln⁡(1+x)x=1\lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1


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:

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:

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


Equivalently, using discrete notation:

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:

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.

Limits Involving Logarithms


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

lim⁡x→0+xln⁡x=0\lim_{x \to 0^+} x \ln x = 0


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.

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:

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:

logarithmic≪polynomial≪exponential\text{logarithmic} \ll \text{polynomial} \ll \text{exponential}


Specifically:

lim⁡x→∞ln⁡xxn=0for any n>0\lim_{x \to \infty} \frac{\ln x}{x^n} = 0 \quad \text{for any } n > 0


lim⁡x→∞xnex=0for any n\lim_{x \to \infty} \frac{x^n}{e^x} = 0 \quad \text{for any } n


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.

Using Special Limits


Rewrite expressions to match known forms.

Example 1


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


Rewrite to expose the standard form:

=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


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


Factor out the coefficient:

=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


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.