Subsections

• Purpose
• Background
  • Simple limits and Maple
  • Limits of more complicated functions
  
  
• Exercises

Limits of functions.

Purpose

The purpose of this lab is to use Maple to become more familiar with limits of functions, including one-sided limits.

Background

Simple limits and Maple

Limits of many functions and expressions can be computed in Maple with the limit command. Some examples are given below.

> limit(x^2+2*x,x=2);
> limit(sin(x)/x,x=0);
> f := x -> (x+3)/(x^2+7*x+12) ;
> limit(f(x),x=-3);
> limit(f(x),x=-4);

If the limit exists, Maple can usually find it. In cases where the limit doesn't exist, Maple gives the answer undefined or sometimes infinity for an unbounded limit or gives a range like -1..1 if the limit doesn't exist, but the expression or function is bounded. See the examples below.

> limit(1/x,x=0);
> limit(sin(1/x),x=0);

You can also use Maple to compute limits as $x$ goes to $\pm \infty$ as shown below.

> f(x);
> limit(f(x),x=infinity);
> limit(f(x),x= -infinity);

Limits of more complicated functions

It should be no secret by now that for most functions $f(x)$ defined by a single formula, $$\lim_{x \rightarrow a} f(x) = f(a)$$ when $f(a)$ exists. For more complicated functions, this may not be true.

If you want to define your own piecewise-defined function, then the Maple piecewise command is the best way to do it. Suppose you wanted to define the following function.

$$g(x) = \left\{ \begin{array}{ll} -x & \mbox{if $x < 0$} \\ x^2+1 & \mbox{if $x \geq 0$} \end{array} \right.$$

Then the Maple command would be the following.

> g := x -> piecewise(x < 0, -x, x^2+1);

If you want to see your function in a more familiar form, just run a command like the one below.

> g(x);

The way the piecewise command works is that you give it a sequence of pairs of conditions and formulas that define your function. When you want to evaluate your function at a particular value of $x$, Maple checks the conditions from left to right until it finds the one that your value of $x$ satisifies. It then plugs the value of $x$ into the next formula. However, notice that the command above only has one condition and two formulas. This is because any value of $x$ is either less than zero or it is greater than or equal to zero, so if a particular value of $x$ fails the first condition, i.e. is not less than zero, it must be greater than or equal to zero and the second formula is the one to use. For more information, see the help page for piecewise.

The limit command works fine for functions that are defined via the piecewise command, as shown in the example below.

> limit(g(x), x=0);
> limit(g(x),x=0, left);
> limit(g(x),x=0, right);
> plot(g(x), x=-0.1..0.1);

Exercises

1. Use Maple to evaluate each of the limits given below.
   1. $$\lim_{x \rightarrow 0} \frac{x}{\sec^2(x)-1}$$
   2. $$\lim_{x \rightarrow \pi} \frac{x}{\sec^2(x)-1}$$
   Although both of the above limits do not exist, the output to each Maple command is different. Use a plot to help you explain the difference between the two limits. Be sure that your plotting range for the $x$ values surround the value that $x$ is approaching and you could also limit your $y$ range to $[-100,100]$ to get a better look at the plots. Then, look at left- and right-handed limits to help explain.

2. For the function $$f(x) = \frac{3x-5+x^2}{\cos(x)}$$
   1. State any points where it is undefined.
   2. Find the limit at $$x = \frac{\pi}{2}$$.
   3. If the limit is undefined, show its right and left-handed limits. Then explain why the limit is undefined.
   4. Plot the function making sure to clearly show the region around the limit point.

3. Given the following peicewise function:
   
   $$f(x) = \left\{ \begin{array}{ll} \cos(x)-1 & \mbox{if $x > 0... ...+\frac{\sin(x)}{x} & \mbox{if $x \leq 0$} \end{array} \right.$$
   
   
   Plot the function. Does $$\lim_{x \rightarrow 0} f(x)$$ exist? To answer this, you must show that the left and right hand limits agree and are finite. Is the function continuous at $x=0$? Explain your reasoning.

4. Given the following peicewise function:
   
   $$f(x) = \left\{ \begin{array}{ll} x \sin(x)+1 & \mbox{if $x < 0$} \\ -x^3+x+1 & \mbox{if $x \geq 0$} \end{array} \right.$$
   
   
   Plot the function. Does $$\lim_{x \rightarrow 0} f(x)$$ exist? To answer this, you must show that the left and right hand limits agree and are finite. Is the function continuous at $x=0$? Explain your reasoning.