Computing Limits I – Limits and Continuity

Posted: 29th November 2012 by seanmathmodelguy in Lectures

Computing  Limits I

In order to compute a limit algebraically, one needs to know what is and more importantly what is not allowed when manipulating limits. For example consider computing\[ \lim_{x\to 1}\frac{x^2-1}{x-1}. \]One cannot just substitute \(x=1\) into this expression because doing so results in the indeterminate form \(0/0\). By factoring the top of the expression one has\[ \lim_{x\to 1}\frac{x^2-1}{x-1} = \lim_{x\to 1}\frac{(x-1)(x+1)}{(x-1)} = \lim_{x\to 1}(x+1) = 2. \]Notice that in the last step one can just plug in \(x=1\) since we get a number and not the indeterminate form \(0/0\). The other indeterminate form is \(\infty/\infty\). As a last example consider\[ \lim_{x\to 0}\frac{x(x+1)}{x^2} = \lim_{x\to 0}\frac{x+1}{x} \to \quad ? \]In this case the limit becomes arbitrarily large and since we do not obtain a single finite value we say that the limit does not exist (DNE). To be precise, \(\lim_{x\to 0^-} x(x+1)/x^2 \to -\infty = L^-\) and \(\lim_{x\to 0^+} x(x+1)/x^2 \to \infty = L^+\) since neither \(L^+\) nor \(L^-\) exists, \(\lim_{x\to 0^-} x(x+1)/x^2\) does not exist. Typically the student neglects to propagate the ‘\(\lim\)’ symbol as one moves from step to step. It is essential that ‘\(\lim\)’ is written at every step otherwise meaningless statements are made. An extreme example of this is while\[ \lim_{x\to 2}(x^2-4) = \lim_{x\to 2}(5x-10) \]since both limits give zero, it is certainly not the case that \(x^2-4 = 5x-10\) for every value of \(x\).

The Limit Laws

The basic algebraic rules regarding the manipulation of limits can be summarized in four statements. Let \(\displaystyle \lim_{x \to a} f(x) = L\) and \(\displaystyle \lim_{x \to a} g(x) = M\) where \(L\) and \(M\) are finite numbers.\[ \begin{array}{llr} \mbox{Sum Rule:} & \displaystyle \lim_{x \to a} [f(x)+g(x)] = L+M, & \hspace{7cm} (1) \\ \mbox{Difference Rule:} & \displaystyle \lim_{x \to a} [f(x)-g(x)] = L-M, & (2) \\ \mbox{Product Rule:} & \displaystyle \lim_{x \to a} f(x)g(x) = L M, & (3) \\ \mbox{Quotient Rule:} & \displaystyle \lim_{x \to a} \left[\frac{f(x)}{g(x)}\right] = \frac{L}{M} \mbox{ provided } M\ne 0. & (4) \end{array} \]One must be careful not to fall into the following trap:\[ 1 = \lim_{x\to 0}1 = \lim_{x \to 0} x \frac{1}{x} = \left( \lim_{x \to 0} x\right) \left( \lim_{x \to 0} \frac{1}{x}\right) = (0)(\infty). \]What happened? Isn’t this just a straight application of the product rule? The violation is in the first line of the limit laws that \(L\) and \(M\) be finite. The moral is that you can break up products in this way only if the individual terms all give finite values. Since, \(\lim_{x\to 0} (1/x)\) is infinite we cannot use these laws. One final rule that may be of some assistance is\[ ~\hspace{7cm}\lim_{x \to a} [f(x)]^{\frac{r}{s}} = L^{\frac{r}{s}} \hspace{6.5cm} (5) \]provided \(r\) and \(s\) are integers and \(s \ne 0\). If \(r/s < 0\), \(L\ne 0\); If \(s\) is even, \(f(x) \ge 0\). We now turn to a few examples.


1. Find \(\displaystyle\lim_{x\to 3}\sqrt{5x+1}\). We evaluate this limit using the limits laws described above. By using the rules (5), (1) and (3) in that order we have\[ \lim_{x\to 3}\sqrt{5x+1} = \sqrt{\lim_{x\to 3}(5x+1)} = \sqrt{5\lim_{x\to 3}x + \lim_{x\to 3}1} = \sqrt{5\cdot 3 + 1)} = 4. \]2. Another direct application of the limit laws gives the following:\[ \lim_{t\to 0}\frac{2t + 3}{6 – 3t} = \frac{\lim_{t\to 0}(2t + 3)}{\lim_{t\to 0}(6 – 3t)} = \frac{2\lim_{t\to 0}t + \lim_{t\to 0}3}{\lim_{t\to 0}6-3\lim_{t\to 0}t} = \frac{0 + 3}{6 – 0} = \frac{1}{2}. \]3. Find \(\displaystyle\lim_{x\to 0}\frac{|x|}{x}\). To evaluate this limit one needs the fact that \(|x| = x\) if \(x \ge 0\) and \(|x| = -x\) if \(x < 0\). Remember the definition of the limit. From the left one has\[ \lim_{x\to 0^-}\frac{|x|}{x} = \lim_{x\to 0^-}\frac{-x}{x} = -1 = L^- \]where we use the fact that \(|x| = -x\) if \(x < 0\).  Approaching \(x = 0\) from the right gives\[ \lim_{x\to 0^+}\frac{|x|}{x} = \lim_{x\to 0^+}\frac{x}{x} = 1 = L^+. \]Since \(L^- \ne L^+\), \(\lim_{x\to 0}|x|/x\) DNE.

Numerical Evaluation of Limits

Since the limit deals with what happens to a function as one approaches a point why can’t one just enter numbers on the calculator that are closer and closer to the desired location where we want the limit?

\(x\) \(f(x)\)
1 0.123105626
0.1 0.124998046
0.01 0.124998000
0.001 0.124000000
0.0001 0
0.00001 0
0.000001 0

The table to the right illustrates what happens when one attempts this procedure to find\[ \lim_{x\to 0} f(x), \hspace{1cm} f(x) = \frac{\sqrt{x^3+16}-4}{x^3}. \]Due to the numerical round off in the calculator the student might mistakenly infer that the limit is zero when in actual fact the exact value is \(1/8\). The algebraic techniques developed in this section and the ones that follow will allow you to determine this exact value analytically. While this procedure does motivate the choice of a particular value for a given limit, it should never be used to ‘prove’ a particular limit. Indeed this numerical evaluation   procedure suffers some serious drawbacks.