Part II · Limits, Continuity, and the Calculus Foundation

Chapter 6 — Powers, Exponentials, Limits, and Continuity

Why This Matters

Calculus relies on understanding what happens near a point, not merely at it. Limits formalise this nearby behaviour. Continuity tells us when small input changes produce small output changes. Exponentials provide the model for growth, decay, and later circuit transients.

6.1 Real powers and exponential functions

For integer exponents, $$a^n$$ means repeated multiplication. For arbitrary real exponents, the clean real-valued theory takes $$a>0$$ . Rational powers use roots, for example $a^{m/n}=\sqrt[n]{a^m}$ when this is real and defined. Irrational exponents are defined consistently through limits of rational approximations.

The functions $$x^a$$ and $$a^x$$ are different. In $$x^a$$ , the base varies; in $$a^x$$ , the exponent varies. For $$a>1$$ , $$a^x$$ passes through $$(0,1)$$ , increases strictly, tends to $$0$$ as $x\to-\infty$ , and grows without bound as $x\to\infty$ . For $$0<a<1$$ , the direction is reversed.

Real powers retain domain conditions. The usual real-valued definition of a^x for arbitrary real exponents begins with a>0. It also distinguishes the variable-base function x^a from the variable-exponent function a^x; their domains and long-run behaviours need not match.

a>0,\qquad a^x=e^{x\ln a}

6.2 Limits of functions

We write

\lim_{x\to c}f(x)=L

when $$f(x)$$ becomes arbitrarily close to $$L$$ as $$x$$ becomes sufficiently close to $$c$$ , without requiring $$x=c$$ . The value $$f(c)$$ may be different from $$L$$ or may not even be defined.

The left and right limits must agree for a two-sided limit to exist:

\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L.

Example — Removable discontinuity

For $x\ne0$ ,

\frac{2^x-1}{x}

is undefined at $$0$$ . Its limit as $x\to0$ is $\ln2$ , a fact derived later from the derivative of $$2^x$$ . Assigning the value $\ln2$ at $$x=0$$ creates a continuous extension. The original expression is not already defined there.

A limit ignores the centre while it is approached. Nearby inputs determine the limiting value, so the function value at the centre may be different or absent. A two-sided limit requires the left and right approaches to agree on one value.

\lim_{x\to c}f(x)=L\Longleftrightarrow\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L

Nearby values can approach a removable hole even when the original expression has no value at the centre.

6.3 Continuity

A function is continuous at $$c$$ when all three conditions hold:

$$f(c)$$ is defined.
$\lim_{x\to c}f(x)$ exists.
The limit equals $$f(c)$$ .

Polynomials are continuous on all real numbers. Rational functions are continuous wherever their denominators are nonzero. Root functions require their real-domain restrictions. These facts are useful because they allow direct substitution only after the domain has been checked.

Continuity requires three linked facts. The point value must exist, the nearby limit must exist, and those two values must agree. Direct substitution is therefore justified for a rational expression only after its denominator restriction has been checked.

f(c)\text{ defined},\quad\lim_{x\to c}f(x)\text{ exists},\quad\lim_{x\to c}f(x)=f(c)

Check that the input is in the domain.
Find the nearby two-sided limit.
Compare it with the actual point value before declaring continuity.

6.4 One-sided limits and discontinuities

A switch can change a circuit model abruptly. At the switching instant, a piecewise function may have a left limit and a different right limit. This is a jump discontinuity. A rational function may have an infinite discontinuity or vertical asymptote where its denominator approaches zero. A missing point with a finite surrounding limit is a removable discontinuity.

A point assignment cannot repair unequal approaches. A jump has distinct left and right limits, whereas a removable discontinuity has one finite surrounding limit and a missing or mismatched point. A vertical asymptote records unbounded nearby behaviour rather than an ordinary finite value.

\lim_{x\to0^-}f(x)=1\ne3=\lim_{x\to0^+}f(x)

Choose a one-sided approach to the declared jump: the left tends to 1 and the right tends to 3.

6.5 Limits at infinity

Limits at infinity describe long-run behaviour. For example,

\lim_{x\to\infty}\frac{3x^2+1}{x^2-4}=3.

The leading powers control the result. Exponentials eventually dominate any fixed polynomial power: for every fixed nonnegative integer $$m$$ ,

\lim_{x\to\infty}x^m e^{-x}=0.

This comparison later explains why exponential transients decay despite polynomial factors that may appear in intermediate calculations.

Divide by the dominant scale. In a rational function, the largest power controls the ratio at large magnitude. Equal degrees give the ratio of leading coefficients; a smaller numerator degree gives zero; a larger numerator degree has no finite horizontal asymptote.

\frac{3x^2+1}{x^2-4}=\frac{3+1/x^2}{1-4/x^2}\longrightarrow3

A rational branch approaches a labelled horizontal asymptote after dominant-power division exposes its long-run value.

Problem-Solving Strategy

Before evaluating a limit, check the domain and try direct substitution. If it produces an indeterminate form such as $$0/0$$ , simplify algebraically by factorisation, rationalisation, or dominant-term division. Use one-sided limits for piecewise definitions and denominator sign changes. Do not perform algebra on the symbol $\infty$ as though it were a real number.

6.6 Limits describe nearby behaviour, not a point calculation

Direct substitution is a useful first test, not the definition of a limit. A $$0/0$$ result says numerator and denominator both become small; it does not say the quotient is zero. Factorisation or rationalisation can expose a simpler nearby expression, provided the excluded input is still recorded.

Example — factor, cancel, then take the limit

\lim_{x\to2}\frac{x^2-4}{x-2} =\lim_{x\to2}\frac{(x-2)(x+2)}{x-2} =\lim_{x\to2}(x+2)=4.

Cancellation is valid only for $x\ne2$ , which is exactly the punctured neighbourhood used by the limit. It does not make the original function defined at $$2$$ .

Example — one-sided limits determine a jump

Let $$f(x)=x+1$$ for $$x<0$$ and $$f(x)=x^2$$ for $x\ge0$ . Then the left limit at $$0$$ is $$1$$ and the right limit is $$0$$ . The two-sided limit does not exist even though $$f(0)$$ is defined; one point value cannot repair unequal approaches.

Example — limits at infinity use dominant terms

\lim_{x\to\infty}\frac{5x^3-2x}{-x^3+7} =\lim_{x\to\infty}\frac{5-2/x^2}{-1+7/x^3}=-5.

Dividing by the highest power exposes the horizontal asymptote. If numerator degree is smaller, the limit is $$0$$ ; if larger, no finite horizontal asymptote results.

[Figure: A removable hole, a jump, and a vertical asymptote; label one-sided limits and a rational graph approaching a horizontal asymptote.]

Cancellation simplifies nearby values, not the original domain. Factoring x²−4 exposes x+2 for x≠2, so the limit is four. The exclusion remains part of the original quotient; the hole is not filled by algebra alone.

\frac{x^2-4}{x-2}=x+2\ (x\ne2),\qquad\lim_{x\to2}\frac{x^2-4}{x-2}=4

Record the input excluded by the original denominator.
Factor and simplify only on the punctured neighbourhood.
Evaluate the simplified nearby expression for the limit without assigning the original function a value there.

Exercises

State the domain of $x^{2/3}$ and of $x^{\sqrt2}$ in the usual real-valued convention.
Evaluate $\lim_{x\to3}(x^2-9)/(x-3)$ and state why substitution initially fails.
Determine the one-sided limits of $$f(x)=|x|/x$$ at $$0$$ .
Find the horizontal asymptote of $$(4x^3-x)/(2x^3+7)$$ .
Sketch a piecewise function with a removable discontinuity at $$x=1$$ and a jump discontinuity at $$x=3$$ .
Evaluate $\lim_{x\to3}(\sqrt{x+1}-2)/(x-3)$ by rationalising.
Give a piecewise rule continuous at $$0$$ but not differentiable there.
Find horizontal asymptotes of $$(2x+1)/(x^2+4)$$ and $$(x^3+1)/(2x^2-1)$$ .