How close can a function get to a value without ever touching it? Limits answer exactly that question, and they are the foundation of everything in calculus: derivatives, integrals, and infinite series all rest on them. In this chapter you will learn to read limits from tables and graphs, compute them with algebra, understand what happens at infinity, and use continuity to guarantee that a function has no surprises.
1. The intuitive idea of a limit
Let \(f\) be a function defined near \(a\) (it may be undefined at \(a\) itself). We write \(\displaystyle\lim_{x\to a} f(x)=L\) when the values \(f(x)\) get as close to \(L\) as we like by taking \(x\) close enough to \(a\), with \(x\neq a\).
The key phrase is “with \(x\neq a\)”. A limit describes the behavior near \(a\), not the value at \(a\). The function can even be undefined at \(a\), as in the next example.
Let \(f(x)=\dfrac{x^2-4}{x-2}\). It is undefined at \(x=2\), but for \(x\neq 2\) we can factor: \(f(x)=\dfrac{(x-2)(x+2)}{x-2}=x+2\). Check numerically:
| \(x\) | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| \(f(x)\) | 3.9 | 3.99 | 3.999 | 4.001 | 4.01 | 4.1 |
The values approach 4 from both sides, so \(\displaystyle\lim_{x\to 2}\dfrac{x^2-4}{x-2}=4\). The graph is the line \(y=x+2\) with one point removed.
2. One-sided limits
Sometimes the function behaves differently on each side of \(a\). We write \(\displaystyle\lim_{x\to a^-}f(x)\) for the left-hand limit (using only \(x
\(\displaystyle\lim_{x\to a}f(x)=L\) if and only if \(\displaystyle\lim_{x\to a^-}f(x)=L\) and \(\displaystyle\lim_{x\to a^+}f(x)=L\). If the two one-sided limits differ, the limit does not exist.
Let \(f(x)=x+1\) for \(x<2\) and \(f(x)=6-x\) for \(x\ge 2\). Then \(\displaystyle\lim_{x\to2^-}f(x)=2+1=3\) and \(\displaystyle\lim_{x\to2^+}f(x)=6-2=4\). Since \(3\neq4\), \(\displaystyle\lim_{x\to2}f(x)\) does not exist, even though \(f(2)=4\).
One-sided limits also describe blow-up. For \(x\) slightly bigger than 0, \(\dfrac1x\) is a huge positive number, so \(\displaystyle\lim_{x\to0^+}\dfrac1x=+\infty\); on the left it is hugely negative, so \(\displaystyle\lim_{x\to0^-}\dfrac1x=-\infty\). The symbol \(\infty\) tells you how the function behaves, but it is not a real number, so such a limit is said to be infinite (and does not exist as a finite number).
3. Limit laws
Suppose \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\) are both finite numbers, and let \(c\) be a constant. Then:
- \(\lim (f+g)=L+M\) and \(\lim (f-g)=L-M\)
- \(\lim\, c\,f=c\,L\) and \(\lim\, f\,g=L\,M\)
- \(\lim \dfrac{f}{g}=\dfrac{L}{M}\) provided \(M\neq0\)
- \(\lim \,(f(x))^n=L^n\) and \(\lim\sqrt[n]{f(x)}=\sqrt[n]{L}\) (when the root is defined)
For polynomials and many other functions, direct substitution works: \(\lim_{x\to a}f(x)=f(a)\). For example, \(\lim_{x\to 3}(2x^2-x)=18-3=15\). Trouble appears when substitution gives the indeterminate form \(\dfrac00\). Then you must rewrite the expression first.
- Substitute \(x=a\). If you get a nonzero number, you are done.
- If you get \(\frac00\), factor the numerator and denominator and cancel the common factor \((x-a)\); or, if a square root is involved, multiply top and bottom by the conjugate.
- Substitute again in the simplified expression.
Find \(\displaystyle\lim_{x\to3}\dfrac{\sqrt{x+1}-2}{x-3}\). Substitution gives \(\frac00\). Multiply by the conjugate \(\sqrt{x+1}+2\):
\[\dfrac{\sqrt{x+1}-2}{x-3}\cdot\dfrac{\sqrt{x+1}+2}{\sqrt{x+1}+2}=\dfrac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}=\dfrac{1}{\sqrt{x+1}+2}.\]
As \(x\to3\), this tends to \(\dfrac{1}{2+2}=\dfrac14\).
4. Limits at infinity and asymptotes
We also ask what happens when \(x\) becomes very large: \(\displaystyle\lim_{x\to+\infty}f(x)\) and \(\displaystyle\lim_{x\to-\infty}f(x)\). The basic fact is \(\displaystyle\lim_{x\to\pm\infty}\dfrac1{x^n}=0\) for every positive integer \(n\).
Divide the numerator and the denominator by the highest power of \(x\) that appears in the denominator, then use \(\frac1{x^n}\to0\).
For \(g(x)=\dfrac{3x+1}{x-1}\), divide by \(x\): \(g(x)=\dfrac{3+\frac1x}{1-\frac1x}\to\dfrac{3}{1}=3\) as \(x\to\pm\infty\). So the line \(y=3\) is a horizontal asymptote. At \(x=1\) the denominator is 0 while the numerator is 4, so \(g\) blows up: \(x=1\) is a vertical asymptote.
The line \(y=L\) is a horizontal asymptote if \(\lim_{x\to+\infty}f(x)=L\) or \(\lim_{x\to-\infty}f(x)=L\). The line \(x=a\) is a vertical asymptote if at least one one-sided limit at \(a\) is \(+\infty\) or \(-\infty\).
5. The squeeze theorem
If \(h(x)\le f(x)\le g(x)\) for all \(x\) near \(a\) (except possibly at \(a\)) and \(\lim_{x\to a}h(x)=\lim_{x\to a}g(x)=L\), then \(\lim_{x\to a}f(x)=L\). The same statement holds as \(x\to\pm\infty\).
Find \(\displaystyle\lim_{x\to0}x^2\sin\dfrac1x\). The function \(\sin\frac1x\) oscillates wildly and has no limit at 0, but it always lies between \(-1\) and \(1\). Multiplying by \(x^2\ge0\):
\[-x^2\le x^2\sin\tfrac1x\le x^2.\]
Both \(-x^2\) and \(x^2\) tend to 0, so the middle function also tends to 0.
6. Continuity at a point
A function \(f\) is continuous at \(a\) if three conditions hold:
- \(f(a)\) is defined;
- \(\lim_{x\to a}f(x)\) exists (as a finite number);
- \(\lim_{x\to a}f(x)=f(a)\).
Informally, you can draw the graph near \(a\) without lifting your pencil. Polynomials, \(\sin x\), \(\cos x\), \(e^x\), and roots are continuous wherever they are defined, and sums, products, quotients (with nonzero denominator), and compositions of continuous functions are continuous.
Let \(f(x)=\dfrac{x^2-9}{x-3}\) for \(x\neq3\). For which value \(k=f(3)\) is \(f\) continuous at 3? We have \(f(x)=x+3\) for \(x\neq3\), so \(\lim_{x\to3}f(x)=6\). Continuity requires \(f(3)=\lim_{x\to3}f(x)\), so \(k=6\).
7. Types of discontinuity
When one of the three conditions fails, \(f\) is discontinuous at \(a\). There are three classic shapes.
- Removable: the limit exists but \(f(a)\) is undefined or different from the limit (a hole). You can repair it by redefining \(f(a)\).
- Jump: both one-sided limits exist but are different numbers.
- Infinite: at least one one-sided limit is \(\pm\infty\) (a vertical asymptote).
8. The Intermediate Value Theorem
If \(f\) is continuous on the closed interval \([a,b]\) and \(N\) is any number between \(f(a)\) and \(f(b)\), then there is at least one \(c\) in \([a,b]\) with \(f(c)=N\).
The theorem is most often used to show that an equation has a solution: if \(f\) is continuous and \(f(a)\) and \(f(b)\) have opposite signs, then \(f\) has a root between \(a\) and \(b\). Continuity is essential; a function with a jump can skip values.
Show that \(x^3+x-3=0\) has a solution in \((1,2)\). Let \(f(x)=x^3+x-3\), a polynomial and therefore continuous. We have \(f(1)=1+1-3=-1<0\) and \(f(2)=8+2-3=7>0\). Since \(0\) lies between \(-1\) and \(7\), the IVT gives some \(c\in(1,2)\) with \(f(c)=0\).
Key takeaways
- \(\lim_{x\to a}f(x)=L\) describes what \(f(x)\) approaches as \(x\) nears \(a\); it ignores \(f(a)\).
- The limit exists exactly when the left-hand and right-hand limits exist and are equal.
- For a \(\frac00\) form, factor and cancel, or use a conjugate, then substitute.
- At infinity, divide by the highest power of \(x\); \(\frac1{x^n}\to0\). Horizontal asymptotes come from limits at \(\pm\infty\), vertical ones from infinite one-sided limits.
- Squeeze theorem: a function trapped between two functions with the same limit has that limit.
- Continuity at \(a\): \(f(a)\) defined, limit exists, and they are equal.
- Discontinuities are removable, jump, or infinite.
- IVT: continuous on \([a,b]\) means every value between \(f(a)\) and \(f(b)\) is reached.
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