Computing Limits I – Limits and Continuity

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.

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.