Limit Definition, Continuity, and Derivatives

Calculus 1, Lectures 8A through 11

The limit definition of the derivative leads to the graph of a function with a hole (a removable discontinuity). The limit still exists, based on the precise epsilon delta definition of a limit. — The graph of the difference quotient $DQ(h)=\frac{(2+h)^{2}-8}{h}$ for computing the derivative $f'(2)$ when $f(x)=x^{3}$ has a removable discontinuity (a hole) at $h=0$ . The limit still exists so that $f'(2)=12$ .

The graph of the difference quotient $DQ(h)=\frac{(2+h)^{2}-8}{h}$ for computing the derivative $f'(2)$ when $f(x)=x^{3}$ has a removable discontinuity (a hole) at $h=0$ . The limit still exists so that $f'(2)=12$ .

Graphs with holes (missing points) might seem like anomalies — like they are mere curiosities hardly worth studying. However, they arise at the heart of differential calculus.

The limit definition of the derivative leads naturally to consideration of a function whose graph has a hole in it. Suppose $f$ is a function defined at and near a number $x=a$ . The derivative of $f$ at $x=a$ is a number written as $f'(a)$ . It is defined by the following limit definition, when it exists:

$f'(a)=\displaystyle\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}.$

Since the number $a$ is fixed, the expression $\frac{f(a+h)-f(a)}{h}$ defines a function of the variable $h$ . Since $\frac{f(a+h)-f(a)}{h}$ is called a difference quotient, it makes sense for us to call this function $DQ$ and write $DQ(h)=\frac{f(a+h)-f(a)}{h}$ .

Because $f$ is defined near the number $x=a$ , it follows that $DQ$ is defined near the number $h=0$ . However, take care to note that $DQ$ is technically undefined at $h=0$ . We say that $DQ$ is defined on a “deleted neighborhood” of $h=0$ .

This does not necessarily mean that the graph of $DQ$ has a hole in it at $h=0$ . But we will see by considering examples that it “typically” does have a hole.

As usual, I start by giving short summaries of some of my calculus lectures at Bethel University in the fall of 2019. After that, I will delve into the main topic of this blog post, which are the typical examples alluded to in the previous paragraph.

Lectures 8A and 8B

I provide a whirlwind tour of various special functions and their graphs in Lectures 8A and 8B. I consider sinusoidal functions, inverse trigonometric functions, polynomial functions, and power functions in Lecture 8A.

Calculus 1, Lecture 8A: Mathematica, Day Length as a Sinusoid, Inverse Trigonometric Functions, Polynomials, Power Functions

Statistical data modeling is a fundamental part of data science. I take data about daylight length in Minneapolis, Minnesota in 2019 and model it with a sinusoidal function. I note that while a sinusoidal function does a pretty good job in modeling the data, it is not completely ideal because the residuals have a pattern. It is more ideal for residuals to be randomly distributed numbers centered on zero.

Next, I review the graphs of inverse trigonometric functions. Finally, I discuss graphs and properties of polynomial and power functions.

Removable Discontinuities (“Holes”) for Rational Functions

The main points of focus in Lecture 8B are power functions and rational functions. In fact, it is in the context of rational functions that I first discuss functions with holes in their graphs. These holes correspond to discontinuities that I describe as “removable”. The phrase “removable discontinuity” does in fact have an official definition.

Calculus 1, Lec 8B: Power Functions, Rational Functions, Asymptotes, Removable Discontinuities, Limits

For rational functions, removable discontinuities arise when the numerator and denominator have common factors which can be completely canceled. For instance, the example I consider in the lecture is $f(x)=\frac{x^{2}-3x-4}{x^{2}-10x+24}$ . I factor the numerator and denominator to get $f(x)=\frac{(x-4)(x+1)}{x-4)(x-6)}$ .

Hence, after factoring, it follows that $f(x)=\frac{x+1}{x-6}$ . But this formula only is true when $x\not=4$ . If $x=4$ , the original formula is undefined. On the other hand, the simplified formula is defined at $x=4$ .

In fact, the simplified function is continuous at $x=4$ . Therefore the original function must have a hole in its graph at $x=4$ . This graph is shown below.

A rational function whose graph has a hole. The limit definition, in other words, the "epsilon-delta" definition of a limit, implies that the limit exists as x approaches 4, in spite of the fact that the function is undefined there. — The graph of $f(x)=\frac{x^{2}-3x-4}{x^{2}-10x+24}=\frac{(x-4)(x+1)}{x-4)(x-6)}$ has a hole at $x=4$ . The reason is that it equals the function $g(x)=\frac{x+1}{x-6}$ when $x\not=4$ . But $g$ is continuous at $x=4$ .

The graph of $f(x)=\frac{x^{2}-3x-4}{x^{2}-10x+24}=\frac{(x-4)(x+1)}{x-4)(x-6)}$ has a hole at $x=4$ . The reason is that it equals the function $g(x)=\frac{x+1}{x-6}$ when $x\not=4$ . But $g$ is continuous at $x=4$ .

What are the coordinates of the missing point? The first coordinate is certainly $x=4$ . The second coordinate is the value of the simplified formula at $x=4$ . This value is $\frac{4+1}{4-6}=-\frac{5}{2}$ . The is the same as the limit of this function as $x$ approaches 4.

The precise definition of a limit, which is discussed in Lectures 9 and 10 below, guarantees the existence of the limit of this function as $x$ approaches 4.

Lectures 9A and 9B

In these lectures, the limit definition is illustrated, first through examples, and then through a precise definition and proof.

Rational Function Examples with Holes and a Jump Discontinuity Example

I look at a three rational function examples with removable discontinuities in Lecture 9A. In addition, I explore the limit of $f(x)=\frac{|x|}{x}$ as $x\rightarrow 0$ .

Calc 1, Lec 9A: Limits of Rational Functions (Asymptotes, Removable Discontinuities), and Abs(x)/x

For each of the rational functions I consider, a common factor of the form $x-c$ is completely canceled. Because of this, in spite of the fact that these functions are undefined at $x=c$ , the limit $\displaystyle\lim_{x\rightarrow c}f(x)$ exists. There are no vertical asymptotes at these points.

Limits as $x\rightarrow \pm \infty$ are also discussed. The key to discovering these limits is based on the relative sizes of the degrees of the numerator (top) and denominator (bottom) of the given rational function.

If the degree of the numerator is larger than the degree of the denominator, then there are no horizontal asymptotes. Polynomial long division must then be done to find any slant asymptotes or curved asymptotes.

If the degree of the numerator is equal to the degree of the denominator, then a horizontal asymptote occurs at $y=k$ , where $k$ is the ratio of the coefficients of the highest powers of $x$ in the numerator and denominator.

Finally, if the degree of the numerator is smaller than the degree of the denominator, then there is a horizontal asymptote at $y=0$ (the $x$ -axis).

On the other hand, the function $f(x)=\frac{|x|}{x}$ has a “jump discontinuity” at $x=0$ . Its output is $+1$ when $x>0$ . But its output is $-1$ when $x<0$ . Therefore, the one-sided limits are different. The right-hand limit is $\displaystyle\lim_{x\rightarrow 1^{+}}f(x)=1$ . But the left-hand limit is $\displaystyle\lim_{x\rightarrow 1^{-}}f(x)=-1$ . Hence, the two-sided limit $\displaystyle\lim_{x\rightarrow 1}f(x)$ does not exist. This is because the two-sided limit will exist if and only if the one-sided limits exist and are equal.

Precise Limit Definition and a Proof

In Lecture 9B, I give the precise (rigorous) definition of a limit and use it in a proof. Before doing these things, I also explore the graph of the “floor” (greatest integer) function $f(x)=\lfloor x\rfloor$ . This function is usually denoted as $\mbox{int}(x)$ on many calculators.

Calc 1, Lec 9B: Limits of the Floor Function, Precise Definition of a Limit, Limit Proof, Continuity

The precise definition of a limit is quite challenging to understand. If you don’t understand it at first, you are in good company. In fact, even Newton and Leibniz did not know about this definition in the late 1600’s and early 1700’s. Neither did the great mathematicians of the rest of the 1700’s. It took until the mid 1800’s for Cauchy and Weierstrass to truly nail this down.

Here it is in its full glory. Let $f$ be a function defined near $x=c$ , but not necessarily at $x=c$ (the graph might have a hole there). We say that $\displaystyle\lim_{x\rightarrow c}f(x)=L$ if the following condition holds. For all numbers $\epsilon>0$ (no matter how small), there exists a number $\delta>0$ (chosen sufficiently small) so that if $0<|x-c|<\delta$ , then $|f(x)-L|<\epsilon$ .

Understanding the Precise Limit Definition

When thinking about this definition, it is essential to draw graphs in various circumstances. It is also essential to realize that for any two numbers $a$ and $b$ , the quantity $|a-b|=|b-a|$ is the distance between $a$ and $b$ along a number line.

I discuss these things in Lecture 9B, along with doing an example proof based on this definition. Take the time to watch it carefully to help you understand it.

In Lectures 10A and 10B, I review the limit definition above and include a Mathematica animation. I also discuss how it is related to continuity and ultimately state the limit definition of the derivative.

The animation is shown below for $f(x)=x^{2}$ , $\epsilon=0.2$ , and $c=1$ . As long as $\delta>0$ is sufficiently small, inputs $x$ between the blue lines will result in outputs $f(x)$ between the red lines. This is equivalent to saying that if $|x-c|=|x-1|<\delta$ , then $|f(x)-L|=|f(x)-1|<\epsilon$ .

Animation to illustrate the precise limit definition (epsilon delta definition of a limit). — Animation to illustrate the limit definition confirming that $\displaystyle\lim_{x\rightarrow 1}x^{2}=1$ . For $\epsilon=0.2$ , we can choose $\delta>0$ sufficiently small so that if $|x-1|<\delta$ , then $|f(x)-1|<\epsilon$ . If $\epsilon$ were even smaller, then $\delta$ could be chosen smaller yet.

Animation to illustrate the limit definition confirming that $\displaystyle\lim_{x\rightarrow 1}x^{2}=1$ . For $\epsilon=0.2$ , we can choose $\delta>0$ sufficiently small so that if $|x-1|<\delta$ , then $|f(x)-1|<\epsilon$ . If $\epsilon$ were even smaller, then $\delta$ could be chosen smaller yet.

Lectures 10A, 10B, and 11

It is reasonably accurate to say that these three lectures form the cornerstones of differential calculus.

Before reviewing the precise definition of a limit in Lecture 10A, I spend some time discussing how rational functions arise in applications. In particular, I review the yield rate of a bond application from Lecture 6A (also see the post “Transformations of Functions”).

Calc 1, Lec 10A: Applications of Rational Functions, Limit Definition, Continuity, Types of Discontinuities

The precise limit definition is then reviewed, along with a Mathematica animation to illustrate it.

Continuous and Discontinuous Functions

Then continuity is explored. Once the definition of a limit has been established, the definition of continuity at a point is easy.

Let $f$ be a function defined both at and near a number $x=c$ . We say that $f$ is continuous at $x=c$ if $\displaystyle\lim_{x\rightarrow c}f(x)=f(c)$ . A function $f$ is then said to be continuous over its domain if it is continuous at every point (number) in its domain.

Ways a Function can be Discontinuous

Next, I explore ways that a function can be discontinuous at a point (number) $x=c$ . There are essentially three ways this can happen, though reason #1 could be considered to be somewhat “phony”.

The function $f$ is undefined at $x=c$ .
The function $f$ is defined at $x=c$ . But $\displaystyle\lim_{x\rightarrow c}f(x)$ does not exist.
The function $f$ is defined at $x=c$ and $\displaystyle\lim_{x\rightarrow c}f(x)$ exists. But $\displaystyle\lim_{x\rightarrow c}f(x)\not=f(c)$ .

Reason #2 typically happens either because of a jump discontinuity or a vertical asymptote.

However, there is a more interesting way that it can happen. A function could oscillate infinitely often with a non-decaying amplitude as $x$ approaches $c$ . The most “famous” example of this is the function $f$ defined by $f(x)=\sin\left(\frac{1}{x}\right)$ when $x\not=0$ and $f(0)=0$ . The graph of this function in a small window near $x=0$ is shown below.

The precise limit definition implies that this function has no limit as x approaches zero because it oscillates infinitely often with constant amplitude. — The graph of $f$ defined by $f(x)=\sin\left(\frac{1}{x}\right)$ when $x\not=0$ and $f(0)=0$ . It oscillates infinitely often with a constant amplitude of 1 over any tiny interval centered on $x=0$ , no matter how small.

The graph of $f$ defined by $f(x)=\sin\left(\frac{1}{x}\right)$ when $x\not=0$ and $f(0)=0$ . It oscillates infinitely often with a constant amplitude of 1 over any tiny interval centered on $x=0$ , no matter how small.

Make sure you take the time to think about why the graph looks this way. See if you can give a rigorous argument for why there are infinitely many oscillations near $x=0$ .

Algebraic Properties of Limits and Continuous Functions and the Limit Definition of the Derivative

In Lecture 10B I talk about useful algebraic properties of limits and continuous functions. I also state the limit definition of the derivative.

Calc 1, Lec 10B: Properties of Limits & Continuous Functions, Limit Definition of the Derivative

Linearity of Limits and Continuity

Two algebraic properties go under the general description of “linearity”. Linearity is often considered to be a property of certain physical systems. However, it has a mathematical description as well.

For example, the following linearity equation is true, assuming all the limits exist. In words, it says that the limit of a linear combination of functions is the corresponding linear combination of their limits.

$\displaystyle\lim_{x\rightarrow c}\left(af(x)+bg(x)\right)=a\cdot \displaystyle\lim_{x\rightarrow c}f(x)+b\cdot \displaystyle\lim_{x\rightarrow c}g(x).$

There is a corresponding linearity property regarding continuous functions, whose truth follows from the limit property above. That property is: if $f$ and $g$ are both continuous at $x=c$ and $a$ and $b$ are constants, then the function $af+bg$ is also continuous at $x=c$ . Note that the function $af+bg$ is defined by the equation $(af+bg)(x)=af(x)+bg(x)$ .

The Limit Definition of the Derivative

The average rate of change of a function $f$ over an interval from $x=a$ to $x=a+h$ is $\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}$ . This represents the slope of the so-called secant line connecting the points $(a,f(a))$ and $(a+h,f(a+h))$ .

As $h$ gets closer and closer to zero, this becomes a rate of change over a smaller and smaller interval. Intuitively, as $h\rightarrow 0$ , it should approach the so-called instantaneous rate of change of $f$ at $x=a$ . This is the value of the derivative of $f$ as $x=a$ , denoted by $f'(a)$ . It also defines the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$ .

We write the following equation, when the limit in question exists. As stated in the introduction, this is the limit definition of the derivative.

$f'(a)=\displaystyle\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$

If $x=a$ is fixed and $h$ is thought of as a variable, then the difference quotient $DQ(h)=\frac{f(a+h)-f(a)}{h}$ is a function of $h$ . Also as stated in the introduction, it is this function whose graph will typically have a hole in it at $h=0$ . Some examples will be explored below.

The Derivative of the Square Root Function and Review for Exam 1

The bulk of Lecture 11 is a review for Exam 1 for my class at Bethel University. Near the beginning, however, I take time to emphasize the superiority of the symbolic approach to derivatives compared to numerical and graphical approximations.

Calc1, Lec 11: Derivative of Sqrt(x) at x = 4 with Graphs & Limit Definition, Review for Exam 1

A Derivative of the Square Root Function

The function in question is $f(x)=\sqrt{x}$ and the point in question is $x=a=4$ . In this case, $DQ(h)=\frac{f(4+h)-f(4)}{h}=\frac{\sqrt{4+h}-2}{h}$ . The derivative we are after is $f'(4)=\displaystyle\lim_{h\rightarrow 0}DQ(h)$ . This limit is, by definition, the slope of the tangent line to the graph of $y=f(x)$ at the point $(a,f(a))=(4,2)$ .

On a “wide” window, the difference quotient function $DQ(h)$ looks fairly “ordinary”, and would seem to have just an ordinary hole in its graph at $h=0$ . From this picture, it seems that $f'(4)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=0.25=\frac{1}{4}$ . This is indeed true, but the second graph below will throw our minds into doubt about this.

The graph of $DQ(h)=\frac{\sqrt{4+h}-2}{h}$ . It seems to have a removable discontinuity at $h=0$ . By the limit definition of the derivative applied to $f(x)=\sqrt{x}$ at $x=a=4$ , it appears that $f'(4)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=0.25=\frac{1}{4}$ .

The graph of $DQ(h)=\frac{\sqrt{4+h}-2}{h}$ . It seems to have a removable discontinuity at $h=0$ . By the limit definition of the derivative applied to $f(x)=\sqrt{x}$ at $x=a=4$ , it appears that $f'(4)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=0.25=\frac{1}{4}$ .

The next graph shows a Mathematica-drawn picture of this function on a window that is very close to zero. Numerical error causes this graph to seem jagged. It is not truly jagged like this seems to imply. Also notice that the outputs are so close to 0.25 that Mathematica is relabeling that number at various points along the vertical axis. Something definitely seems wrong here.

A zoomed in version of the graph of $DQ(h)=\frac{\sqrt{4+h}-2}{h}$ . Now it seems that $\displaystyle\lim_{h\rightarrow 0}DQ(h)$ does not exist. This is a false impression, however. This graph is inaccurate because of numerical computing error.

The Symbolic Approach is Superior

A symbolic approach to proving that $f'(4)=\frac{1}{4}$ dispels the confusion caused by the graph above. However, the computation involves a trick that is not obvious. We must rationalize the numerator of the fraction defining $DQ(h)$ .

Here is the calculation. The equality of the expressions $\frac{\sqrt{4+h}-2}{h}$ and $\frac{1}{\sqrt{4+h}+2}$ when $h\not=0$ is required to say the limits are the same (assuming they exist). The continuity of the expression $\frac{1}{\sqrt{4+h}+2}$ at $h=0$ is required to do the final substitution.

$f'(4)=\displaystyle\lim_{h\rightarrow 0}\frac{\sqrt{4+h}-2}{h}\cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\displaystyle\lim_{h\rightarrow 0}\frac{4+h-4}{h(\sqrt{4+h}+2)}$

$=\displaystyle\lim_{h\rightarrow 0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{2+2}=\frac{1}{4}$

More Examples

Let’s end now by considering a couple more examples to illustrate the main point. The main point is that $DQ(h)$ usually has a hole in its graph at $h=0$ . In other words, it has a removable discontinuity because $\displaystyle\lim_{h\rightarrow 0}DQ(h)$ exists. This is equivalent to the derivative $f'(a)$ existing.

In these examples, nothing too strange goes on when technology is used to graph $DQ(h)$ .

A Derivative of the Cubing Function

Let $f(x)=x^{3}$ . What is $f'(2)$ ? Let’s first compute and simplify $DQ(h)$ . Here is the relevant calculation.

$DQ(h)=\frac{f(2+h)-f(2)}{h}=\frac{(2+h)^{3}-2^{3}}{h}=\frac{2^{3}+3\cdot 2^{2}h+3\cdot 2h^{2}+h^{3}-8}{h}$

$=\frac{h(12+6h+h^{2})}{h}=12+6h+h^{2}$

The last equality is true as long as $h\not=0$ . Therefore, as functions of $h$ , the graph of $DQ(h)$ is the same as the graph of $12+6h+h^{2}$ except when $h=0$ .

Hence, the graph of $DQ(h)$ is a parabola with a point missing, as shown below.

The graph of $DQ(h)=\frac{f(2+h)-f(2)}{h}$ when $f(x)=x^{3}$ . From this graph, we see that $f'(2)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=12$ , in spite of the fact that $DQ(h)$ is undefined at $h=0$ .

The graph of $DQ(h)=\frac{f(2+h)-f(2)}{h}$ when $f(x)=x^{3}$ . From this graph, we see that $f'(2)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=12$ , in spite of the fact that $DQ(h)$ is undefined at $h=0$ .

The derivative in question is $f'(2)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=\displaystyle\lim_{h\rightarrow 0}12+6h+h^{2}=12+6\cdot 0+0^{2}=12$ .

A Derivative of the Reciprocal Function

Let $f(x)=\frac{1}{x}$ . What is $f'(2)$ ? Let’s first compute and simplify $DQ(h)$ . Here is the relevant calculation.

$DQ(h)=\frac{f(2+h)-f(2)}{h}=\frac{1}{h}\cdot \left(\frac{1}{2+h}-\frac{1}{2}\right)=\frac{1}{h}\cdot \left(\frac{2-(2+h)}{2(2+h)}\right)=\frac{-h}{2h(2+h)}=-\frac{1}{4+2h}$

The last equality is true as long as $h\not=0$ . Therefore, as functions of $h$ , the graph of $DQ(h)$ is the same as the graph of $-\frac{1}{4+2h}$ except when $h=0$ .

Hence, the graph of $DQ(h)$ is the one shown below with a point missing.

The graph of $DQ(h)=\frac{f(2+h)-f(2)}{h}$ when $f(x)=\frac{1}{x}$ . From this graph, we see that $f'(2)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=-\frac{1}{4}$ , in spite of the fact that $DQ(h)$ is undefined at $h=0$ .

The derivative in question is

$f'(2)=\displaystyle\lim_{h\rightarrow 0}DQ(h)=\displaystyle\lim_{h\rightarrow 0}\left(-\frac{1}{4+2h}\right)=-\frac{1}{4+2\cdot 0}=-\frac{1}{4}$ .