Limit laws and algebraic methods
Combine limits you know, then rewrite expressions that resist substitution.
The idea
Limit laws let us calculate with approached values . Start with substitution. When the relevant laws apply, it gives the limit; when substitution produces \(0/0\), look for an equivalent expression at nearby inputs.
Theorem
Limit laws
Suppose \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\) are finite, and \(c\) is constant. Then
\[\begin{aligned}
\lim_{x\to a}(f(x)\pm g(x))&=L\pm M,\\
\lim_{x\to a}c f(x)&=cL,\\
\lim_{x\to a}f(x)g(x)&=LM,\\
\lim_{x\to a}\frac{f(x)}{g(x)}&=\frac{L}{M},\qquad M\ne0.
\end{aligned}\]
For integers \(n\ge2\), the root law is
\[\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}.\]
For even roots, require \(f(x)\ge0\) near the point and \(L\ge0\). These laws also apply to one-sided limits on the relevant side.
Example 1
Check the denominator first
\[\lim_{x\to-1}\frac{2x^2+3}{x+4}
=\frac{2(-1)^2+3}{-1+4}=\frac53.\]
The denominator approaches \(3\ne0\), so the quotient law applies.
Rewriting near the point
Principle
Equal nearby expressions have the same limit
If \(f(x)=g(x)\) for every input sufficiently close to \(a\) except possibly \(a\) itself, they have the same limit whenever either limit exists. Their values at \(a\) may differ or be missing.
For the graph below, factoring gives
\[q(x)=\frac{x^3-8}{x-2}=x^2+2x+4=p(x),\qquad x\ne2.\]
The graphs agree at every input except \(x=2\), where \(q\) is undefined and \(p(2)=12\). Both limits as \(x\to2\) equal \(12\).
Prerequisite · algebra
Conjugate identity
The difference-of-squares identity gives, for \(A,B\ge0\),
\[(\sqrt{A}-\sqrt{B})(\sqrt{A}+\sqrt{B})=(\sqrt{A})^2-(\sqrt{B})^2=A-B.\]
The two factors are conjugates .
Worked examples
Example 2
Factor before cancelling
Substitution in \(\lim_{x\to-2}\frac{x^2+5x+6}{x+2}\) gives \(0/0\). For \(x\ne-2\),
\[\frac{x^2+5x+6}{x+2}=\frac{(x+2)(x+3)}{x+2}=x+3.\]
Therefore,
\[\lim_{x\to-2}\frac{x^2+5x+6}{x+2}=\lim_{x\to-2}(x+3)=1.\]
The cancellation does not assign a value to the original quotient at \(-2\).
Example 3
Remove a difference of square roots
To find \(\lim_{x\to5}\frac{\sqrt{x+4}-3}{x-5}\), multiply by the conjugate. Near \(5\), with \(x\ne5\),
\[\begin{aligned}
\frac{\sqrt{x+4}-3}{x-5}
&=\frac{\sqrt{x+4}-3}{x-5}\cdot\frac{\sqrt{x+4}+3}{\sqrt{x+4}+3}\\
&=\frac{(\sqrt{x+4})^2-3^2}{(x-5)(\sqrt{x+4}+3)}\\
&=\frac{x-5}{(x-5)(\sqrt{x+4}+3)}\\
&=\frac{1}{\sqrt{x+4}+3}.
\end{aligned}\]
Now the denominator approaches a nonzero value:
\[\lim_{x\to5}\frac{\sqrt{x+4}-3}{x-5}
=\lim_{x\to5}\frac{1}{\sqrt{x+4}+3}=\frac16.\]
Example 4
Combine fractions first
Find \(\lim_{h\to0}\frac{\frac{1}{2+h}-\frac12}{h}\). For nearby \(h\ne0\),
\[\frac{\frac{1}{2+h}-\frac12}{h}
=\frac{2-(2+h)}{2(2+h)h}=-\frac{1}{2(2+h)}.\]
Hence
\[\lim_{h\to0}\frac{\frac{1}{2+h}-\frac12}{h}
=\lim_{h\to0}\left(-\frac{1}{2(2+h)}\right)=-\frac14.\]
Example 5
Use the expression on each side
Let \(r(x)=\frac{\lvert x\rvert}{x}\) for \(x\ne0\). Recall that \(\lvert x\rvert=-x\) when \(x<0\), and \(\lvert x\rvert=x\) when \(x>0\). Thus
\[\lim_{x\to0^-}\frac{\lvert x\rvert}{x}=\lim_{x\to0^-}(-1)=-1,\qquad
\lim_{x\to0^+}\frac{\lvert x\rvert}{x}=\lim_{x\to0^+}1=1.\]
The one-sided limits disagree, so the two-sided limit does not exist. Algebraic simplification must respect the side of approach.
Warning
The quotient law cannot divide by a zero limiting denominator. In particular, \(0/0\) is an indeterminate form , not a value: it leaves the limit undecided. Simplify or examine the two sides before drawing a conclusion. A factor may approach zero and still be cancelled at nearby inputs: for example, \(x-5\ne0\) when \(x\ne5\), so cancelling it is valid while taking a limit as \(x\to5\).
Quiz
Enable JavaScript for interactive answer checking.
As \(x\to a\), suppose \(f(x)\to-2\) and \(g(x)\to3\). What is \(\displaystyle\lim_{x\to a}\frac{f(x)^2+g(x)}{g(x)-1}\)?
\(1/2\) \(7/2\) \(-1/2\) The laws do not apply Explanation The denominator approaches \(2\ne0\), so the quotient law applies: \(((-2)^2+3)/(3-1)=7/2\).A square plate has side length \(3+h\). What is \(\displaystyle\lim_{h\to0}\frac{(3+h)^2-9}{h}\), the limiting change in area per change in side length?
Explanation Expand the numerator to \(6h+h^2\). For \(h\ne0\), the quotient is \(6+h\), so its limit is \(6\). Both dimensions of the square change.In \(\displaystyle\lim_{x\to0}\frac{\sqrt{4+x}-2}{x}\), which expression equals the quotient for nearby \(x\ne0\)?
\(\sqrt{4+x}+2\) \(1/\sqrt{4+x}\) \(1/(\sqrt{4+x}-2)\) \(1/(\sqrt{4+x}+2)\) Explanation Multiplying by the conjugate makes the numerator \((4+x)-4=x\), which cancels the denominator. The rewritten limit is \(1/4\).Suppose \(f(x)\to0\) and \(g(x)\to0\) as \(x\to0\). Which claim about \(f(x)/g(x)\) is justified by these facts alone?
Its limit is \(1\) They do not determine the quotient limit Its limit is \(0\) It cannot have a finite limit Explanation The quotient law requires a nonzero denominator limit. Taking \(f(x)=cx\), \(g(x)=x\) gives any chosen finite limit \(c\); taking \(f(x)=\lvert x\rvert\), \(g(x)=x\) gives no two-sided limit.A calibration rule is \(C(t)=\frac{t^2-25}{t-5}\) for \(t<5\), and \(C(t)=b-t\) for \(t>5\). Which \(b\) makes \(\displaystyle\lim_{t\to5}C(t)\) exist?
Explanation The left branch equals \(t+5\) for \(t\ne5\), giving left limit \(10\). The right limit is \(b-5\); equality requires \(b=15\). No assigned value at \(t=5\) is needed for the limit.For \(x\ne3\), a function equals \(\frac{x^2-2x-3}{x-3}\), but its assigned value is \(F(3)=-100\). Find \(\displaystyle\lim_{x\to3}(F(x)-4)\).
\(0\) \(-104\) \(4\) The limit does not exist Explanation Away from \(3\), factor to get \(F(x)=x+1\). Thus \(\lim_{x\to3}(F(x)-4)=\lim_{x\to3}(x-3)=0\). The isolated value \(F(3)\) does not affect this limit.
Course contents