Continuity – Limits and Continuity

Posted: 12th February 2013 by seanmathmodelguy in Lectures


A function \(f(x)\) is said to be continuous at a point \(a\) in its domain if the following three properties hold.

  • \(\displaystyle \lim_{x \to a} f(x)\) exists. This takes three steps to show in itself.
  • \(f(a)\) has to exist,
  • \(\displaystyle \lim_{x \to a} f(x) = f(a)\).

Continuity connects the behaviour of a function in a neighbourhood of a point with the value at the point.

If the domain of the function is bounded say \([a,b]\) then each end point of the interval can only be approached in one way. The left end point is at \(x = a\) so \(f(x)\) is said to be continuous at the left end point ‘\(a\)’ if \(\displaystyle \lim_{x \to a^+} f(x) = f(a)\). In a similar fashion, \(f(x)\) is said to be continuous at the right end point ‘\(b\)’ if \(\displaystyle \lim_{x \to b^-} f(x) = f(b)\).

Types of Discontinuities

Since there are only a few ways that the limit of a function cannot exist at a point there are few ways that a function can fail to be continuous. These are classified into four types.
1. Jump Discontinuity (also known as a simple discontinuity)
2. Removable Discontinuity
3. Infinite Discontinuity
4. Oscillatory Discontinuity

An explicit example of each type of discontinuity follows next.


1. Jump Discontinuity:
Is \(
f(x) = \begin{cases}
\displaystyle x-1, & 1 \le x \le 2 \\
-1, & -2 \le x < 1
\) continuous at \(x = 1\)? Where is \(f(x)\) continuous?

To answer this we go back to the definition. By computing \(L^+ = 0\) and \(L^- = -1\) (for \(x = 1\)) we see that they are not equal and consequently, \(\displaystyle \lim_{x \to 1} f(x)\) DNE. Therefore \(f(x)\) is not continuous at \(x=1\). As to where \(f(x)\) is continuous, this is everywhere else in the domain \((-2,1) \cup (1,2)\). For the endpoints, in this case we say that \(f(x)\) is continuous from the right at \(x=-2\) and continuous from the left at \(x=2\).
A graph of the function appears to the right.m0601b

2. Removable Discontinuity:
Consider \(
g(x) = \begin{cases}
\displaystyle x, & x \ne 2 \\
5, & x = 2.
\) Is \(g(x)\) continuous at \(x = 2\)? No because even though \(\displaystyle \lim_{x\to 2}g(x) = 2\) exists, \(\displaystyle \lim_{x\to 2}g(x) \ne g(2) = 5\). Since this function can be made continuous by redefining it at \(x=2\), we call this type of discontinuity removable.

To illustrate how a function can be fixed, notice that \(
f_1(x) = \displaystyle\frac{\sin x}{x}
\) is not continuous for all \(x\in {\Bbb R}\) since \(f_1(0)\) DNE. However, \(
f_2(x) = \begin{cases}
\displaystyle \frac{\sin x}{x}, & x \ne 0 \\[3mm] 1, & x = 0
\) is continuous for all \(x\in {\Bbb R}\).

3. Infinite Discontinuity:m0601d
Consider the function \(y = 1/x\) on any domain that includes \(x=0\). Since the function becomes unbounded continuity fails at \(x=0\) since the limit does not exist there. The domain of the function is very important since the same function on a different domain (\(\displaystyle y = 1/x, 1 \le x \le 2\)), does not have an infinite discontinuity because it does not become unbounded on the given domain.

4. Oscillatory Discontinuity:
This type of discontinuity occurs when a function oscillates too much, as in the case of \(y = \sin(1/x)\). As \(x\to 0\), \(f(x)\) does not approach a single value.

m0601eWe leave it as an exercise to the student to show that
\( f(x) = \begin{cases}
\displaystyle x\sin\left(\frac{1}{x}\right), & x \ne 0 \\[3mm] 0, & x = 0
\) is a continuous function for all \(x\in {\Bbb R}\).

Strategy to Calculate Limits – Limits and Continuity

Posted: 9th February 2013 by seanmathmodelguy in Lectures

Strategy to Calculate Limits


To compute \(\displaystyle \lim_{x \to a} f(x)\):

A. Try to plug the value of \(a\) directly into the function.

    • If we get a number or the limit ‘blows up’ then we are done!
    • You should be so lucky. Typically the value is undefined, having the form \(\displaystyle \frac{0}{0}\) or \(\displaystyle \frac{\infty}{\infty}\).

B. If we do not get a number then we need to simplify the expression.

    • Use the definition of the limit;
    • Use of the limit rules;
    • Factoring;
    • Multiplying by the conjugate;
    • Finding a common denominator;
    • Using the Squeeze Theorem;
    • Applying some memorized limit such as \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1.\)

C. Try to plug the number into the function once again. If we get a number or \(\pm \infty\) then we are done. Otherwise go back to step B.

A quick word of warning…. l’Hôpital’s theorem should not be used at this point since it involves taking derivatives. Most instructors don’t give any credit for limits found using this method at this point.

Like anything else, the best way to get proficient at finding limits is with practice. We conclude with a few examples.


1. Find \(\displaystyle \lim_{x\to 1} f(x)\) where \(f(x) = \displaystyle \begin{cases}x, & x < 1 \\ 0, & x = 1 \\ -x+2, & x > 1.

This one requires the definition of the limit. From the left one has\[
\lim_{x\to 1^-}f(x) = \lim_{x\to 1^-} x = 1 = L^-
\]and from the right,\[
\lim_{x\to 1^+}f(x) = \lim_{x\to 1^+} (-x+2) = 1 = L^+
\]since \(L^+=L^- = 1\), \(\displaystyle \lim_{x\to 1} f(x)=1\).

2. Find \(\displaystyle \lim_{x\to -1} \frac{\sqrt{x^2+8}}{2x+4}\).

For this one we just plug in to find \(\sqrt{9}/2 = 3/2\).

3. Find \(\displaystyle \lim_{x\to -\infty} \frac{x^4+1}{x^3-1}\).

If we try to plug in we get the indeterminate form \(\infty/\infty\). The trick here is to divide the top and bottom by the highest power in the denominator.\[
\lim_{x\to -\infty} \frac{x^4+1}{x^3-1} =
\lim_{x\to -\infty} \frac{x^4+1}{x^3-1} \frac{(1/x^3)}{(1/x^3)} =
\lim_{x\to -\infty} \frac{x+1/x^3}{1-1/x^3} \to -\infty
\]so the limit does not exist.

4. Find \(\displaystyle \lim_{x\to 4} \frac{x-4}{\sqrt{x}-2}\).

This limit requires that we multiply by the conjugate.\[
\lim_{x\to 4} \frac{x-4}{\sqrt{x}-2} \frac{(\sqrt{x}+2)}{(\sqrt{x}+2)} =
\lim_{x\to 4} \frac{(x-4)(\sqrt{x}+2)}{x-4} =
\lim_{x\to 4} \frac{\sqrt{x}+2}{1} = 2+2 = 4
\]allowing us to conclude that the limit is 4.

5. Find \(\displaystyle \lim_{x\to 0} \frac{\sec x – 1}{x^2}\).

This one is a bit of a challenge. We start with multiplying by the conjugate, then use the trigonometric identity \(\sec^2 x – 1 = \tan^2 x\) and finally we separate into three pieces.
\lim_{x\to 0} \frac{\sec x – 1}{x^2} &=
\lim_{x\to 0} \frac{\sec x – 1}{x^2}
\frac{(\sec x + 1)}{(\sec x + 1)} =
\lim_{x\to 0} \frac{\tan^2 x}{x^2(\sec x + 1)} \\ &=
\lim_{x\to 0} \left(\frac{\sin x}{x}\right)^2
\lim_{x\to 0} \left(\frac{1}{\cos^2 x}\right)
\lim_{x\to 0} \left(\frac{1}{\sec x+1}\right) \\ &=
1^2 \cdot \frac{1}{1} \cdot \frac{1}{2} = \frac{1}{2}.

Limits Involving Infinity – Limits and Continuity

Posted: 9th February 2013 by seanmathmodelguy in Lectures

Limits Involving Infinity

Let’s start with what we mean when we say \(\displaystyle \lim_{x \to\infty} f(x) = L\) or \(\displaystyle \lim_{x \to-\infty} f(x) = L\).m0401a

We say \(f(x)\) has limit \(L\) as \(x\) approaches infinity (\(\infty\)) and write \(\displaystyle \lim_{x \to\infty} f(x) = L\) if as \(x\) moves increasingly far from the origin in the positive direction, \(f\) gets arbitrarily close to \(L\).

In an analogous fashion, we say \(f(x)\) has limit \(L\) as \(x\) approaches minus infinity (\(-\infty\)) and write \(\displaystyle \lim_{x \to -\infty} f(x) = L\) if as \(x\) moves increasingly from the origin in the negative direction, \(f\) gets arbitrarily close to \(L\).

Horizontal and Vertical Asymptotes

There are essentially two ways in which infinity can be involved with a given function \(f(x)\). We can either ask how the function behaves if \(x\to\pm\infty\) or find values of \(x\) where the function itself becomes arbitrarily large. The former question deals with finding horizontal asymptotes while the latter question deals with vertical asymptotes.

  • We say the function $f(x)\) has a horizontal (like the horizon) asymptote \(y = L\) if either \(\displaystyle \lim_{x \to \infty} f(x) = L\) or \(\displaystyle \lim_{x \to -\infty} f(x) = L\).
  • We say the function \(f(x)\) has a vertical asymptote at \(x = a\) if either \(\displaystyle \lim_{x \to a^+} f(x) = \pm \infty\) or \(\displaystyle \lim_{x \to a^-} f(x) = \pm \infty\).

1. Determine the asymptotes of the function
f(x) = \left\{ \begin{array}{ll}
\displaystyle \frac{1}{(x-1)^2}, & x \ge 0 \\
\displaystyle \frac{x+1}{x}, & x < 0.
\end{array} \right.
\] This example is little tricky in the sense that the function is defined piecewise and as a result there are two separate horizontal asymptotes.m0401b
Focusing on the branch for \(x \ge 0\), there is a vertical asymptote at \(x=1\) and since \(\displaystyle \lim_{x\to\infty}f(x) = 0\) there is a horizontal asymptote of \(y = 0\).

For the other branch, \(\displaystyle \lim_{x\to -\infty}f(x) = 1\) giving a horizontal asymptote of \(y = 1\). There is one final vertical asymptote at \(x=0\) since \(\displaystyle \lim_{x\to 0^-}f(x) = -\infty\). The final graph is to the right. Note that the horizontal asymptote \(y=0\) is shown slightly above the line \(y=0\) for clarity.

Oblique Asymptotes

An oblique asymptote describes the situation when the limiting behaviour of the function is a line. One particular way that this can happen is with a rational polynomial \(f(x) = P(x)/Q(x)\) with the \(\mbox{deg }P = \mbox{deg } Q+1\). In this case the asymptote is neither horizontal nor vertical but rather occurs at an angle.

Determining the horizontal, vertical and oblique asymptotes places constraints on the behaviour of the function. This will be illustrated in the example that follow.

2. Determine the asymptotes of the function \(\displaystyle f(x) = \frac{4-x^{2}}{x+1}\). First of all, there are no horizontal asymptotes since \(\displaystyle \lim_{x\to\pm\infty}f(x) = \mp\infty\). There is a vertical asymptote at \(x = -1\) since \(f(x)\) becomes arbitrarily large at this value. In addition we have an oblique asymptote of \(y = -x+1\) which is found using long division since
\frac{4-x^{2}}{x+1} = -x + 1 + \frac{3}{x+1}.

Computing Limits II: The Squeeze Theorem – Application Proof

Posted: 9th February 2013 by seanmathmodelguy in Lectures

One very important application of the squeeze theorem is the proof that \(\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\). We present this proof next.

This proof is for the limit as \(\theta\to 0^+\). The case for \(\theta\to 0^-\) can be proved in exactly the same manner and we leave it as an exercise to the student.
Consider the figure to the right. Notice that the area of the triangle OAP is less than the area of the sector OAP which is in turn less than the area of the triangle OAT. Let’s estimate each of these areas in turn.
\mbox{Area of triangle OAP} &=
\frac{1}{2} \mbox{base} \times \mbox{height}\\
&= \frac{1}{2} (1) \sin \theta = \frac {1}{2} \sin \theta.\\
\mbox{Area of sector OAP} &=
\frac{1}{2} r^2 \theta = \frac{1}{2} (1) \theta =
\frac {1}{2} \theta.\\
\mbox{Area of triangle OAT} &=
\frac{1}{2} (1) \tan\theta = \frac{1}{2} \tan\theta.
Using our initial observation, one has $$\frac{1}{2} \sin \theta \le \frac{1}{2} \theta \le \frac{1}{2} \tan \theta, \hspace{1cm} 0 \le \theta < \frac{\pi}{2}.\hspace{2cm}(1)$$The inequality for \(-\pi/2 < \theta \le 0\) is \(\tan\theta\le\theta\le\sin\theta\). From the left hand side of (1) one has \(\sin\theta\le\theta\) or \((\sin\theta)/\theta\le 1\) if \(\theta \ge 0\). From the right hand side we have \(\theta \le \tan\theta\) or \(\cos\theta\le (\sin\theta)/\theta\) if \(\theta \ge 0\). In other words,$$\cos \theta \le \frac{\sin \theta}{\theta} \le 1$$provided \(0\le\theta<\pi/2\). Again, we leave it to the student to verify that this last inequality is true for \(-\pi/2<\theta\le 0\). At this point, we look at the limit as \(\theta\to 0\) and apply the squeeze theorem:$$\lim_{\theta\to 0} \cos \theta \le
\lim_{\theta\to 0} \frac{\sin \theta}{\theta} \le \lim_{\theta\to 0} 1.$$Since the left and right hand sides has a limit of one, we can conclude that $$\lim_{\theta\to 0} \frac{\sin \theta}{\theta} = 1.$$This limit is used so often in calculus it is a good idea to remember it. We leave this material with two examples that use this new result.

1. Find \(\displaystyle \lim_{x\to 0}\frac{\sin^2 x}{x}\).

Again, we use the limit laws, $$\lim_{x\to 0} \frac{\sin^2 x}{x} = \left(\lim_{x\to 0} \frac{\sin x}{x}\right) (\lim_{x\to 0} \sin x) = 1\cdot 0 = 0.$$

2. Find \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}\).
For this one we need to multiply by the conjugate of the numerator,
\lim_{x\to 0}\frac{1-\cos x}{x^2} &=
\lim_{x\to 0}\frac{1-\cos x}{x^2}
\left(\frac{1+\cos x}{1+\cos x}\right) =
\lim_{x\to 0}\frac{1-\cos^2 x}{x^2(1+\cos x)}\\ &=
\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)} =
\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2
\lim_{x\to 0}\frac{1}{1+\cos x}
= 1^2 \cdot \frac{1}{2} = \frac{1}{2}.