The limit laws

Statement of the limit laws and first examples

  • Suppose that \(c\) is a constant, and the limits \(\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x)\) exist. Then we have the following formulas.
    • (The reason for the strange numbering is that I wanted them to agree with the numbering of the textbook.)
      1-2. Sum and Difference law. \(\lim_{x\to a}(f(x)\pm g(x))=(\lim_{x\to a}f(x))\pm(\lim_{x\to a}g(x))\)
    1. Product law. \(\lim_{x\to a}f(x)g(x)=(\lim_{x\to a}f(x))(\lim_{x\to a}g(x))\)
    2. Quotient law. If \(\lim_{x\to a}g(x)\ne0\), then we have \(\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\).
    3. \(\lim_{x\to a}c=c\)
    4. \(\lim_{x\to a}x=a\).
    5. Constant multiple law. Rules 4 and 7 give \(\lim_{x\to a}(cf(x))=c\lim_{x\to a}f(x)\).
    6. Power law. Let \(n\) be a positive integer. Applying the Product law repeatedly for \(g(x)=f(x)\), we get \(\lim_{x\to a}(f(x))^n=(\lim_{x\to a}f(x))^a\).
    7. Root law. Let \(n\) be a positive integer. In case \(n\) is even, let's assume that \(\lim_{x\to a}f(x)>0\). Then we have \(\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]\lim_{x\to a}f(x)\).
    • Exercises. 2.3: 4,8
    • Direct substitution property. If \(f(x)\) is a polynomial or a rational function, and \(a\) is in the domain of \(f(x)\), then we have \(\lim_{x\to a}f(x)=f(a)\).
    • All the limit laws have one-sided versions.

More examples

  • Recall that we have guessed \(\lim_{x\to1}\frac{x-1}{x^2-1}=\frac12\).
    • Note that we can't use direct substitution because \(f(x)=\frac{x-1}{x^2-1}\) is not defined at \(x=1\)
    • We can't use the Quotient law either, because the denominator \(x-1\) has limit \(0\) as \(x\to1\).
    • But we can use that when calculating this limit, we only need to consider values \(x\) near \(1\) and not equal to \(1\).
    • Therefore, from the fraction \(\frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}\) we can cancel the common factor \((x-1)\) inside the limit: \[ \lim_{x\to1}\frac{x-1}{x^2-1}=\lim_{x\to1}\frac{x-1}{(x-1)(x+1)}=\lim_{x\to1}\frac{1}{x+1}=\frac12. \]
  • The same rule holds in general: Suppose that the limits \(\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x)\) exist, and that we have \(f(x)=g(x)\) when \(x\ne a\). Then we have \(\lim_{x\to a}f(x)=\lim_{x\to a}g(x)\).
    • Let \[ g(x)=\begin{cases} x+1& x\ne1,\\ \pi & x=1. \end{cases} \]
    • Since \(g(x)=x+1\) whenever \(x\ne1\), we have \[ \lim_{x\to1}g(x)=\lim_{x\to1}(x+1)=2. \]
    • Exercises. 2.3: 12,20,28,32.


  • We can now calculate this limit, which was trouble when we tried to do this numberically.
    • The trick is to get rid of the square root in the numerator: \[ \frac{\sqrt{t^2+9}-3}{t^2}=\frac{\sqrt{t^2+9}-3}{t^2}\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}=\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}=\frac{1}{\sqrt{t^2+9}+3}. \]
    • With this, we can calculate the limit: \[ \lim_{t\to0}\frac{\sqrt{t^2+9}-3}{t^2}=\lim_{t\to0}\frac{1}{\sqrt{t^2+9}+3}=\frac16. \]
    • Exercises. 2.3: 22,30,34,56,61-63