Posted: 25th January 2013 by seanmathmodelguy in Blog

I’ve received the following email from ICIAM and wanted to bring this to the wider mathematics community.

Dear Colleagues,

The January 2013 issue — Volume 1, No 1 — of the ICIAM Newsletter is now available.

Please visit www.iciam.org/news to download a PDF copy of the Newsletter from the link that you will find there, and to subscribe if you would like notices of future issues.

Best regards,
Barbara Keyfitz

##### Contents, Volume 1, Number 1:

2 What ICIAM Does for Its Members — Barbara Lee Keyfitz
4 Mathematics of Planet Earth 2013 (MPE2013) — Christiane Rousseau
5 CIM International Conferences and Advanced Schools Planet Earth, Portugal 2013
5 News from the EMS — Franco Brezzi
7 ICIAM and ICSU — Roger Pfister & Barbara Keyfitz
8 About the Code of Practice of the European Mathematical Society — Arne Jensen
9 Mathematical Congress of the Americas, 2013 — Susan Friedlander & Marcelo Viana
10 Two Items from SEMA
11 Start-up of the “Desk for Mathematics in Industry” in Italy — Roberto Natalini
11 IMA Bulletin — Iain Duff
12 ICIAM in the News
13 Inside the SMAI — Maria J. Esteban
14 ICIAM 2015: Call for Proposals of Thematic and Industrial Minisymposia
14 Call for nominations for ICIAM Prizes for 2015

## 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.

###### 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.

## Introduction to Limits – Limits and Continuity

Posted: 13th October 2012 by seanmathmodelguy in Lectures

#### Introduction to Limits

The limit of a function is concerned with the behaviour of a function near a given point. What happens at the point is of no concern. To be more precise,

if “the limit of the function $$f(x)$$ as $$x$$ approaches the point $$a$$” is the value $$L$$ then this is denoted as$\lim_{x\to a}f(x) = L.$

Consider the graph to the right where$f(x) = \left\{ \begin{array}{ll} 3, & x \ne 4 \\ 5, & x = 4. \end{array} \right.$What is $$\displaystyle\lim_{x\to 4}f(x)$$? There are two ways to approach $$x=4$$. From the left and from the right. From the left the limit is $$3$$ and from the right the limit is $$3$$. Since these limits from the left and right are the same, we can say that$\lim_{x\to 4}f(x) = 3.$

Note that the limit has nothing to do with what happens at $$x=4$$ since $$f(4) = 5$$ which is not equal to $$\displaystyle\lim_{x\to 4}f(x) = 3$$.

###### Graphical Way of Looking at Limits

Now consider the graph to the left. What is $$\displaystyle\lim_{x \to 3} f(x)$$? One can approach this point two ways. From the left, $$\displaystyle\lim_{x \to 3^-} f(x) = 1$$ and from the right $$\displaystyle\lim_{x \to 3^+} f(x) = 4$$.

In this example the limit does not exist (DNE) because the limit we get as we approach $$x=3$$ depends on the direction in which we approach $$x=3$$.

Because of this problem of obtaining a different limit based on the direction in which we approach a point we need the concept of left and right hand limits.

###### Right and Left Hand Limits

Notice that $$\lim_{x \to 0} f(x)$$ DNE. Therefore, the limit you get depends on how you approach $$x = 0$$.

As you approach from the left (also called from below’) $$\lim_{x \to 0^-} f(x) = 4 = L^-$$.

As you approach from the right (also called from above’) $$\displaystyle \lim_{x \to 0^+} f(x) = -3 = L^+$$.

In order for the limit to exist at a point, we first require that both the left and right hand limits must exist. By this we mean that both $$L^+$$ and $$L^-$$ are finite numbers. Second, we require that these values are the same, $$L^+ = L^-$$.

We can summarize these ideas in the following theorem.
Theorem 1.1 We say that the limit of $$f(x)$$ as $$x$$ approaches $$a$$ exists and equals $$L$$, denoted as $$\displaystyle\lim_{x\to a} f(x) = L$$
1. if $$\displaystyle \lim_{x \to a^-} f(x) = L^-$$ exists,
2. $$\displaystyle \lim_{x \to a^+} f(x) = L^+$$ exists,
3. and $$L^+ = L^- = L$$.

Note that both $$L^+$$ and $$L^-$$ must exist and not be $$\pm\infty$$.

## Computing Limits II: The Squeeze Theorem – Limits and Continuity

Posted: 27th August 2012 by admin in Lectures

#### The Squeeze Theorem

Theorem 1.1  Suppose $$g(x) \le f(x) \le h(x)$$ and this is true for all $$x$$ in a neighbourhood of $$x = a$$ (except perhaps at $$x = a$$). Further, suppose $$\displaystyle \lim_{x \to a} g(x) = \displaystyle \lim_{x \to a} h(x) = L$$. Then $$\displaystyle \lim_{x \to a} f(x) = L$$.

The idea here is that $$f(x)$$is trapped (or squeezed) between$$g(x)$$ and $$h(x)$$.

Sometimes the squeeze theorem is referred to as the sandwich theorem since the limit is ‘sandwiched’ between two values.
This idea of squeezing the function between two other functions is illustrated in the following example.

Examples:

1. Evaluate $$\displaystyle \lim_{x \to 0} x^{2}f(x)$$, where $$-3 \le f(x) \le 4$$. Notice that $$-3x^2 \le x^2f(x) \le 4x^2$$ so that the limit,$\lim_{x \to 0} -3x^2 \le \lim_{x \to 0} x^2f(x) \le \lim_{x \to 0} 4x^2.$Since the limit of both the left and right hand side equal zero, the function $$x^2f(x)$$ is squeezed between zero and zero. Therefore the limit is zero!
2. Determine $$\displaystyle \lim_{x \to 0} x^4\left[6 + \sin\left(\frac{3}{x^2}\right)\right]$$.
Knowing $$-1 \le \sin x \le 1$$, one has $$-1 \le \sin\displaystyle\left(\frac{3}{x^2}\right) \le 1$$ so that $$5 \le 6 + \sin\left(\frac{3}{x^{2}}\right) \le 7$$ or $$5x^4 \le x^4 \left[6 + \sin\left(\frac{3}{x^{2}}\right)\right] \le 7x^4.$$Taking the limit,$$\displaystyle \lim_{x \to 0}5x^4 \le \lim_{x \to 0}x^4\left[6 + \sin\left(\frac{3}{x^{2}}\right)\right] \le \lim_{x \to 0}7x^4.$$The limit of the left hand and right hand side are both zero, the initial function is squeezed in between zero and zero, forcing its limit to be zero as well.
3. Find $$\displaystyle \lim_{x\to\infty}\frac{\cos x}{x}$$.
Since $$-1\le \cos x\le 1$$, we have for $$x > 0$$ that$$-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}.$$By taking limits one has$$-\lim_{x\to\infty} \frac{1}{x} \le \lim_{x\to\infty} \frac{\cos x}{x} \le \lim_{x\to\infty} \frac{1}{x}.$$The limits of both the left and right hand sides is zero, so we have effectively squeezed the limit between zero and zero. Therefore, the desired limit must be zero!