Category Archives: Limits

Fun with Continuity

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Most functions we see in calculus are continuous everywhere or at all but a few points that can be easily identified. But consider the Dirichlet function:

Q\left( x \right)=\left\{ \begin{matrix} 1 & x\in \ rational\ numbers \\ 0 & x\in \ irrational\ numbers \\ \end{matrix} \right.

Since there is one (actually many) rational numbers between any two irrational number and one (many again) irrational numbers between any two rational numbers, this function is not continuous anywhere!

But a very similar function is continuous at exactly one point (1, 1):

f\left( x \right)=\left\{ \begin{matrix} 1 & x\in \ rational\ numbers \\ x & x\in \ irrational\ numbers \\ \end{matrix} \right.

Can you use this idea to make a function that is continuous at exactly two points?

Asymptotes

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Horizontal asymptotes are the graphical manifestation of limits as x approaches infinity. Vertical asymptotes are the graphical manifestation of limits equal to infinity (at a finite x-value).

Thus, since \displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\left( 1-{{2}^{-x}} \right)=1. The graph will show a horizontal asymptote at y = 1.

Since the graph of \displaystyle y={{2}^{-x}}\sin \left( x \right) approaches the x-axis as an asymptote, it follows that \underset{x\to \infty }{\mathop{\lim }}\,\left( {{2}^{-x}}\sin \left( x \right) \right)=0. (The fact that this graph crosses the x-axis many times on its trip to infinity is not a concern; the axis is still an asymptote.)

Since \underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{2}}}=\infty ,\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{1}{x}=-\infty ,\text{ and }\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{x}=\infty , the functions all have a vertical asymptote of x = 0.

 

 

 

 

Continuity

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Limits logically come before continuity since the definition of continuity requires using limits. But practically and historically, continuity comes first. The concept of a limit is used to explain the various kinds of discontinuities and asymptotes. Start by studying discontinuities.

Types of discontinuities to consider: removable (a gap or hole in the graph), jump, infinite (vertical asymptotes), oscillating, and end behavior (horizontal asymptotes).

Numerically: Make a table for the value of \displaystyle \frac{1}{x-3} near x = 3 and as \displaystyle x\to \pm \infty . Relate the values and their signs to the graph. (Divide by a small number get a big number; divide by a big number, get a small number.)

Use the vocabulary of limits to explain the features of graphs. Example: The function \displaystyle \frac{{{x}^{2}}-4}{x-2} has no value at x = 2 (f(2) does not exist), but as you get closer to x = 2 the function value gets closer to 4 (\displaystyle \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=4).

Relate the limit, value and graph of the function. In the example above, the graph looks like the line y=x+2 with a gap or hole at the point (2, 4). Another example: \displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3 since, the graph gets closer to y = 3 as you go farther to the right. The line y = 3 is a horizontal asymptote.

Do this numerically as well:  \displaystyle \frac{3{{x}^{2}}+x+8}{{{x}^{2}}+1}=3+\frac{x+5}{{{x}^{2}}+1} and since the fraction gets smaller as |x| gets larger, the function approaches 3 from above when x > 0 and from below when x < 0 (why?)

Extra for your experts: Discuss the reason for the jump discontinuity of

\displaystyle f\left( x \right)=\frac{\cos \left( x \right)\sqrt{{{x}^{2}}-2x+1}}{x-1}

Dominance

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When considering functions made up of the sums, differences, products or quotients of different sorts of functions (polynomials, exponentials and logarithms), or different powers of the same sort of function we  say that one function dominates the other. This means that as x approaches infinity or negative infinity, the graph will eventually look like the dominating function.

  • Exponentials dominate polynomials,
  • Polynomials dominate logarithms,
  • Among exponentials, larger bases dominate smaller,
  • Among polynomials, higher powers dominate lower,

For example, consider the function x{{e}^{x}}. The exponential function dominates the polynomial. As x\to \infty , the graph looks like an exponential approaching infinity; that is, \displaystyle \underset{x\to \infty }{\mathop{\lim }}\,x{{e}^{x}}=\infty . As  x\to -\infty the graph looks like an exponential with very small negative values (i.e. small in absolute value); so, \displaystyle \underset{x\to -\,\infty }{\mathop{\lim }}\,x{{e}^{x}}=0.

Another example, consider a rational function (the quotient of two polynomials). If the numerator is of higher degree than the denominator, as x\to \pm \infty the numerator dominates, and the limit is infinite. If the denominator is of higher degree, the denominator dominates, and the limit is zero. (And if they are of the same degree, then the limit is the ratio of the leading coefficients. Dominance does not apply.)

Dominance works in other ways as well. Consider the graphs of y=3{{x}^{2}} and y={{2}^{x}}. In a standard graphing window, the graphs appear to intersect twice. But on the right side the exponential function is lower than the polynomial. Look farther out and farther up, the exponential dominates and will eventually lie above the polynomial   (after x = 7.334).

Here’s an example that pretty much has to be done using the dominance approach.

\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\frac{\ln \left( {{x}^{5}} \right)}{{{x}^{0.02}}}=0

The polynomial function in the denominator, even with the very small exponent, will dominate the logarithm function. The denominator will eventually get larger than the numerator and drive the quotient towards zero. We will return to this function when we know about finding maximums and points of inflection and find where it starts decreasing. For more on this see my post Far Out!

Finding Limits

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Ways to find limits (summary):

  1. If the function is continuous at the value x approaches, then substitute that value and the number you get will be the limit.
  2. If the function is not continuous at the value x approaches, then
    1. If you get something that is not zero divided by zero, the limit does not exist (DNE) or equals infinity (see below).
    2. If you get \frac{0}{0} or \frac{\infty }{\infty } the limit may exist. Simplify by factoring, or using different trig functions. Later in the year a method called L’Hôpital’s Rule can often be used in this situation.
  3. Dominance is a quick way of finding many limits. Exponentials dominate, polynomials, polynomials dominate logarithms, higher powers dominate lower powers. The next post will give some hints about dominance.

Infinity is not a number, but it often is used as if it were. When we say a limit is infinity, what we mean is that the function increases without bound, or there is some x-value that will make the expression larger than any number you choose. Writing things like \infty -\infty =0,\frac{\infty }{\infty }=1,\infty +\infty =2\infty  are common mistakes.

DNE or Infinity?  \displaystyle \underset{x\to 3}{\mathop{\lim }}\,\frac{1}{{{\left( x-3 \right)}^{2}}} does not exist, and DNE is a correct answer. However, it is a bit better to say the limit is (equals) infinity, indicating that the expression gets larger without bound as x approaches 3. Both answers will get credit on an AP exam.  \displaystyle \underset{x\to 3}{\mathop{\lim }}\,\frac{1}{x-3} DNE since the one-sided limits (from the left and from the right) are different.  Only DNE gets credit here.

Take a look at this AP question 1998 AB-2: In (a) students found that \displaystyle \underset{x\to \infty }{\mathop{\lim }}\,2x{{e}^{2x}}=\infty \text{ or }DNE, in (b) they found the minimum value of 2x{{e}^{2x}}  is -{{e}^{-1}} and in (c) they had to state the range of the function is [-{{e}^{-1}},\infty )\text{ or }x>-{{e}^{-1}}. Thus making the students show they knew that this kind of DNE is the kind where the value increases without bound.

Deltas and Epsilons

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The Formal Definition of Limit

Teachers are often concerned that the Advanced Placement Calculus tests do not test the formal definition of limit and the delta-epsilon idea. I think there are two reasons for this.

  1. For linear functions the relationship is always \delta =\frac{\varepsilon }{\left| m \right|} where m is the slope of the line. Anyone can memorize that, so testing it does not provide any evidence that the student knows anything. For any other function (polynomials of higher degree, trigonometric functions, etc.) some special one-time “trick” is required, often different for each type of function. Thus, the question becomes unreasonably difficult even for one who understands the concept.
  2. As a multiple choice question, since “none of these” is never a choice on AP tests, the smallest answer must be correct. If \delta =\tfrac{1}{2}\varepsilon  is correct, then so is \delta =\tfrac{1}{3}\varepsilon ,\ \delta =\tfrac{1}{10}\varepsilon  etc. Again no real knowledge of the definition is required – just choose the smallest delta.

Of course, the definition of limit is important. You may teach it in your class. It just will not appear on the AP Calculus exams.

Here are some notes The Definition of Limit contributed by Paul A . Foerster of San Antonio, Texas, that will help your students understand the concept.

Why Limits?

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There are four important things before calculus and in beginning calculus for which we need the concept of limit.

    1. The first is continuity. Most of the time in pre-calculus mathematics and in the calculus we deal with nice continuous functions or functions that are not continuous at just a few points. Limits give us the vocabulary and the mathematics necessary to describe and deal with discontinuities of functions. Historically, the modern (delta-epsilon) definition of limit comes out of Weierstrass’ definition of continuity.
    2. Asymptotes: A vertical asymptote is the graphical feature of function at a point where its limit equals positive or negative infinity. A horizontal asymptote is the (finite) limit of a function as x approaches positive or negative infinity.

Ideally, one would hope that students have seen these phenomena and have used the terms limit and continuity informally before they study calculus. This is where the study of calculus starts. The next two items are studied in calculus and are based heavily on limit.

3. The tangent line problem. The definition of the derivative as the limit of the slope of a secant line to a graph is the first of the two basic ideas of the calculus. This single idea is the basis for all the concepts and applications of differential calculus.

4. The area problem. Using limits it is possible to find the area of a region with a curved side, even if the curve is not something simple like a semi-circle. The definite integral is defined as the limit of a Riemann sum and gives the area regions with a curved side. This then can be extended to huge number of very practical applications many having nothing to do with area.

So these are the main ways that limits are used in beginning calculus. Students need a good visual  understanding, what the graph looks like,  of the first two situations listed above and how limits describe and define them. This is also necessary later when third and fourth come up.