From Continuity to an Explicit Radius
To prove continuity at a point, the epsilon-delta definition asks us to control the output whenever the input is sufficiently close. For polynomials and rational functions, the algebraic form of the function lets us choose that radius explicitly. The essential steps are to bound powers near the input point and, for a rational function, to ensure its denominator stays away from zero.
We use the standard metric on \(\mathbb{R}\), so the distance between \(x\) and \(a\) is \(|x-a|\). A map \(f:D\to\mathbb{R}\), where \(D\subseteq\mathbb{R}\), is continuous at \(a\in D\) if for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(x\in D\) and \(|x-a|<\delta\) imply \(|f(x)-f(a)|<\varepsilon\). The chosen \(\delta\) may depend on \(a\), on \(\varepsilon\), and on the coefficients of the function.
A Local Estimate for Polynomials
Let \(p(x)=\sum_{j=0}^{m}c_jx^j\) be a real polynomial, and fix \(a\in\mathbb{R}\). Restricting \(|x-a|<1\) gives \(|x|<|a|+1\). This places both \(x\) and \(a\) in a bounded region, where each difference of powers has a simple bound.
Proof. For every integer \(j\ge1\), the factorization \(x^j-a^j=(x-a)(x^{j-1}+x^{j-2}a+\cdots+a^{j-1})\) has \(j\) terms in parentheses. Since \(|x-a|<1\), we have \(|x|<|a|+1=M\), while \(|a|\le M\). Each term in the parentheses therefore has absolute value at most \(M^{j-1}\), so \(|x^j-a^j|\le jM^{j-1}|x-a|\). Now \(p(x)-p(a)=\sum_{j=1}^{m}c_j(x^j-a^j)\). The triangle inequality gives \(|p(x)-p(a)|\le\sum_{j=1}^{m}|c_j|jM^{j-1}|x-a|=L_p|x-a|\), as required. \(\square\)
The constant term contributes nothing to a difference. The estimate also covers constant polynomials: then the sum defining \(L_p\) is empty and equals zero, and \(p(x)-p(a)=0\).
Proof. If \(L_p=0\), the estimate in the proposition gives \(|p(x)-p(a)|\le0\) whenever \(|x-a|<1\), so the choice \(\delta=1\) works for every \(\varepsilon>0\). If \(L_p>0\) and \(|x-a|<\delta\), then \(|x-a|<1\), so the proposition applies. Also \(\delta\le\varepsilon/L_p\); hence \(|p(x)-p(a)|\le L_p|x-a|<L_p\delta\le\varepsilon\). This proves the required strict inequality. \(\square\)
Worked Example: A Cubic at \(a=1\)
Find an explicit radius that proves continuity of \(p(x)=2x^3-3x+1\) at \(a=1\) for \(\varepsilon=1/10\).
Here \(M=|1|+1=2\). The nonconstant coefficients are \(c_1=-3\) and \(c_3=2\), so the estimate constant is \(L_p=1\cdot|-3|\cdot2^0+3\cdot|2|\cdot2^2=3+24=27\). Also \(p(1)=2-3+1=0\). Since \(L_p>0\), take \(\delta=\min(1,(1/10)/27)=\min(1,1/270)=1/270\).
If \(|x-1|<1/270\), the polynomial estimate gives \(|p(x)-p(1)|\le27|x-1|<27/270=1/10\). Thus this explicit \(\delta\) works. The estimate is not necessarily the largest possible radius; it is a radius whose validity follows directly from the proved bound.
Check: The factorization also gives \(p(x)-p(1)=(x-1)(2x^2+2x-1)\). For \(|x-1|<1/270<1\), we have \(|x|<2\), so \(|2x^2+2x-1|\le2|x|^2+2|x|+1<8+4+1=13\). Consequently \(|p(x)-p(1)|<13/270<27/270=1/10\), confirming the claimed tolerance with a direct check.
Worked Example: A Quadratic at \(a=-2\)
For \(p(x)=x^2\), find a radius for \(\varepsilon=1/5\) at \(a=-2\). Here \(M=|-2|+1=3\), and the sole nonzero nonconstant coefficient is \(c_2=1\). Thus \(L_p=2\cdot1\cdot3^1=6\). Since \(p(-2)=(-2)^2=4\), the formula gives \(\delta=\min(1,(1/5)/6)=\min(1,1/30)=1/30\).
Whenever \(|x-(-2)|=|x+2|<1/30\), the estimate applies and yields \(|x^2-4|\le6|x+2|<6/30=1/5\). Therefore the selected radius proves the required continuity statement at \(-2\).
Check: The exact factorization is \(x^2-4=(x+2)(x-2)\). From \(|x+2|<1/30<1\) we get \(|x|<3\), and hence \(|x-2|\le|x|+2<5\). It follows that \(|x^2-4|<5/30=1/6<1/5\), which independently verifies the requested tolerance.
Rational Maps: Protect the Denominator First
For a rational map \(r(x)=p(x)/q(x)\), its natural domain is \(D=\{x\in\mathbb{R}:q(x)\ne0\}\). A continuity estimate at \(a\in D\) must control not only the numerator’s change but also the denominator. The condition \(q(a)\ne0\) supplies a positive margin: if \(q(x)\) changes by less than \(|q(a)|/2\), then \(|q(x)|\) remains at least \(|q(a)|/2\).
Proof. First suppose \(|x-a|<\delta_q\). In particular, \(|x-a|<1\), so the polynomial estimate for \(q\) applies. If \(L_q=0\), it gives \(|q(x)-q(a)|\le0\). If \(L_q>0\), the definition of \(\delta_q\) gives \(|q(x)-q(a)|\le L_q|x-a|<L_q\delta_q\le|q(a)|/2\). In either case \(|q(x)-q(a)|\le|q(a)|/2\). Therefore the reverse triangle inequality gives \(|q(x)|\ge|q(a)|-|q(x)-q(a)|\ge|q(a)|/2>0\). In particular, \(x\in D\).
For such \(x\), subtract the fractions and split the numerator change from the denominator change: \(\left|\frac{p(x)}{q(x)}-\frac{p(a)}{q(a)}\right|\le\frac{|p(x)-p(a)|}{|q(x)|}+\frac{|p(a)|\,|q(a)-q(x)|}{|q(x)|\,|q(a)|}\). The polynomial estimates give \(|p(x)-p(a)|\le L_p|x-a|\) and \(|q(x)-q(a)|\le L_q|x-a|\). Using \(|q(x)|\ge|q(a)|/2\) in the two denominators yields \(\left|r(x)-r(a)\right|\le\left(\frac{2L_p}{|q(a)|}+\frac{2|p(a)|L_q}{|q(a)|^2}\right)|x-a|=C|x-a|\). If \(C=0\), this bound gives \(|r(x)-r(a)|=0<\varepsilon\). If \(C>0\) and \(|x-a|<\delta\), then \(|r(x)-r(a)|\le C|x-a|<C\delta\le\varepsilon\). This proves both that the map is defined nearby and that it satisfies the epsilon-delta condition. \(\square\)
Worked Example: A Rational Map at \(a=0\)
Find an explicit radius for \(r(x)=(x^2+1)/(x+2)\) at \(a=0\) for \(\varepsilon=1/10\). The point is in the domain because \(q(0)=0+2=2\ne0\), and \(r(0)=(0^2+1)/(0+2)=1/2\).
Here \(M=|0|+1=1\). For \(p(x)=x^2+1\), \(L_p=2\cdot1\cdot1^1=2\); for \(q(x)=x+2\), \(L_q=1\cdot1\cdot1^0=1\). The denominator radius is \(\delta_q=\min(1,|2|/(2\cdot1))=\min(1,1)=1\). The combined constant is \(C=2\cdot2/2+2\cdot|1|\cdot1/|2|^2=2+2/4=5/2\). Thus \(\delta=\min(1,(1/10)/(5/2))=\min(1,1/25)=1/25\).
If \(|x|<1/25\), then \(|r(x)-r(0)|\le(5/2)|x|<(5/2)(1/25)=1/10\). The denominator estimate also verifies that \(x\) is in the domain: \(|q(x)-q(0)|=|x|<1/25<1=|q(0)|/2\), so \(q(x)\ne0\).
Check: Direct subtraction gives \(r(x)-r(0)=\frac{x^2+1}{x+2}-\frac12=\frac{2x^2-x}{2(x+2)}=\frac{x(2x-1)}{2(x+2)}\). For \(|x|<1/25\), \(|2x-1|\le2|x|+1<27/25\), and \(|x+2|\ge2-|x|>49/25\). Hence \(|r(x)-r(0)|<\frac{(1/25)(27/25)}{2(49/25)}=27/2450<1/10\); the last inequality follows from \(270<2450\). This direct calculation confirms the requested tolerance.
Worked Example: A Point Near a Pole
Consider \(r(x)=(x+1)/(x-1/4)\) at \(a=0\), with \(\varepsilon=1\). Since \(q(0)=0-1/4=-1/4\ne0\), the point belongs to the domain, and \(r(0)=1/(-1/4)=-4\).
Again \(M=|0|+1=1\). For \(p(x)=x+1\), \(L_p=1\cdot1\cdot1^0=1\); for \(q(x)=x-1/4\), \(L_q=1\cdot1\cdot1^0=1\). Therefore \(\delta_q=\min(1,|{-1/4}|/(2\cdot1))=\min(1,1/8)=1/8\). The rational estimate constant is \(C=2\cdot1/(1/4)+2\cdot|1|\cdot1/(1/4)^2=8+32=40\). Thus \(\delta=\min(1/8,1/40)=1/40\).
For \(|x|<1/40\), the denominator changes by \(|q(x)-q(0)|=|x|<1/40<1/8=|q(0)|/2\). It remains nonzero, and the rational estimate gives \(|r(x)-r(0)|\le40|x|<40/40=1\). The denominator is near zero at the chosen point, so the denominator-control step is essential; the resulting radius is small but valid.
Check: Direct subtraction gives \(r(x)-r(0)=\frac{x+1}{x-1/4}+4=\frac{5x}{x-1/4}\). Since \(|x|<1/40\), the reverse triangle inequality gives \(|x-1/4|\ge1/4-|x|>1/4-1/40=9/40\). Thus \(|r(x)-r(0)|<\frac{5/40}{9/40}=5/9<1\). This verifies the same tolerance and confirms that the denominator cannot vanish within the chosen radius.
A Common Error and a Useful Strategy
Common mistake. For \(p(x)=x^2\) at \(a=2\), an incorrect estimate is \(|x^2-4|=|x-2|^2\). It omits the factor \(x+2\): the correct factorization is \(|x^2-4|=|x-2||x+2|\). The incorrect estimate would suggest that \(\delta=1\) works for \(\varepsilon=1\), since \(|x-2|<1\) would appear to give \(|x^2-4|<1\). But \(x=5/2\) satisfies \(|x-2|=1/2<1\), while \(|x^2-4|=|25/4-16/4|=9/4>1\). The corrected local bound uses \(|x+2|\le|x|+2\), and when \(|x-2|<1\), \(|x|<3\), so \(|x+2|<5\). Hence \(|x^2-4|<5|x-2|\); for example \(\delta=1/5\) ensures \(|x^2-4|<1\).
The general method is to choose a simple local restriction first, usually \(|x-a|<1\), then bound each factor there. For a rational map, first spend part of the available radius to keep the denominator away from zero, and only then control the difference of the fractions. These bounds are designed for reliability rather than sharpness: smaller valid radii are often easier to calculate and justify.
Check Your Understanding
For each question, use the local estimates and state the domain condition when the function is rational.
- For \(p(x)=3x^2-2x+5\) at \(a=1\), calculate \(M\) and \(L_p\), then find the stated bound using \(\varepsilon=1/10\).
- For \(p(x)=x^3\) at \(a=0\), apply the polynomial estimate with \(\varepsilon=1/8\). What radius does it give?
- For \(r(x)=(x+2)/(x+3)\) at \(a=0\), calculate \(L_p,L_q,\delta_q\), and \(C\), then choose a radius for \(\varepsilon=1/5\).
- Why does the rational-map estimate require \(q(a)\ne0\)? Identify where that assumption is used in the proof.
- For a constant polynomial \(p(x)=c\), explain why the choice \(\delta=1\) works for every \(\varepsilon>0\).