# Topic 2 Limits

## 2.1 Understand limits using tangent lines

Intuitively, the limit of a function \(f\) at \(x=a\) is a fixed value \(L\) that values of the function \(f(x)\) approach as the values of \(x\) \((\ne a)\) approach \(a\).

The slope of a tangent line to the graph of a functions is a limit of slopes of secant lines. This can be visualized easily using Maple with the support of the package `Student[Calculus1]`

.

Like predefined functions in Maple, package consists of predefine commands for Maple. The package `Student`

serves for studying Calculus and other subject interactively. The subpackage `Student[Calculus1]`

focus mainly on Calculus as the name indicated.

As different package has different focus and serves for different purpose, Maple won’t load a specific package until you run the command `with(package_name)`

. For example, the command
`with(Student[Calculus1])`

will load the subpackage `Calculus1`

.

**Example 2.1 **
Observe how do secant lines of the function \(f(x)=x^3-2\) approach to the tangent line at \(x=1\).

*Solution. * Load the `Student[Calculus1]`

package using `with()`

.

`with(Student[Calculus1])`

Use `TangentSecantTutor`

from the loaded package to observe changes of secant lines.

`TangentSecantTutor(x^3-1, x=1)`

**Exercise 2.1 **Explore the package `Student`

, in particular the subpackage `Student[Calculus1]`

.
You can use the command `?Student`

to get help.

`TangentSecantTutor`

command.
## 2.2 Estimate limits numerically or graphically

To estimate a limit \(\lim\limits_{x\to a}f(x)\) numerically, one may pick some values close to \(a\) and evaluate the function. In Maple, the calculation can be done by using the repetition statement `for counter in array do statement end to;`

.

**Example 2.2**Estimate the limit \(\lim\limits_{x\to 0}\frac{\sin x}{x}\) by approximations.

*Solution. *
First, we pick some values close to 0, for example -0.01, -0.001, -0.0001, 0.01, 0.001, 0.0001 and assign them to an expression.

`sq:=[-0.01, -0.001, -0.0001, 0.0001, 0.001, 0.01]`

Now we find the function values using two new commands instead of defining the function a priori.

`for t in sq do evalf(subs(x=t, sin(x)/x)) end do;`

Graphs provide visual intuition which helps understand and solve problems. Recall, the command `plot(expression, domain, option)`

produces a graph of the function defined by the expression over your choice of domain.

**Example 2.3 **
Determine whether the limit \(\lim\limits_{x\to 0}\frac{1}{1- \cos x}\) exists.

*Solution. *
Apply the `plot`

function to the expression over the domain \((-0.5, 0.5)\).

`plot(1/(1-cos(x)), x=-0.5..0.5)`

The graph shows that the function \(y=\frac{1}{1- \cos x}\) goes to \(\infty\) when \(x\) approaches \(0\). So the limit is an infinite limit.

**Exercise 2.2 **
Estimate the limit \(\lim\limits_{t \to 0}\frac{1-\cos x}{x}\) numerically.

**Exercise 2.3 **
Determine whether the limit \(\lim\limits_{x \to 1}\frac{\sin x}{|x-1|}\) exists using the graph.

## 2.3 Evaluate limits

Maple provides the following command to evaluate a limit

```
limit(function, position, direction)
```

The direction may be omitted when evaluating a two-side limit.

**Example 2.4 **
Determine whether the limit \(\lim\limits_{x\to 0}\frac{|x|}{x}\) exists.

*Solution. *
You may find the left and right limits using the following commands.

```
limit(abs(x)/x, x=0, left);
limit(abs(x)/x, x=0, right);
```

It turns out that \(\lim\limits_{x\to 0}\frac{|x|}{x}\) does not exist because the left limit and the right limit are different.

**Example 2.5 **
Evaluate the limit \(\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\), where \(f(x)=\dfrac{1}{x}\).

*Solution. *
The limit can be obtained using the following command.

`limit((1/(x+h)-1/x)/h, h=0)`

**Exercise 2.4 **
Determine whether the limit \(\lim\limits_{x\to 1/2}\dfrac{2x-1}{|2x^3-x^2|}\) exists.

**Exercise 2.5 **
Evaluate the limit \(\lim\limits_{t\to 0}\dfrac{\sqrt{x+t}-\sqrt{x}}{t}\).

## 2.4 Learn limit laws using `LimitTutor`

Suppose the limits of two functions \(f\) and \(g\) at the same point \(x=a\) exist (equal finite numbers). Then the limit operation commutes with addition/subtraction, multiplication/division and power.

In Maple, you may use the command `LimitTutor(function, position, direction)`

, which is again supported by the subpackage `Student[Calculus1]`

, to learn how to evaluate a limit using limit laws and theorems.

**Example 2.6 **
Evaluate \(\lim\limits_{x\to 0}\dfrac{1-\cos x}{x}\).

*Solution. *
Load the subpackage `Student[Calculus1]`

if it was not loaded.

`with(Student[Calculus1])`

Use the following command to learn how to evaluate the limit.

`LimitTutor(2-x^3, x=2,'right')`

You will see an interactive windows pop out. You can choose the see the procedure step-by-step.

**Exercise 2.6 **
Evaluate \(\lim\limits_{x\to 0}\dfrac{(x+2)(\cos x-1)}{x^2-x}\) using `LimitTutor`

.

## 2.5 Squeeze Theorem

Comparison is a very useful tool in problem solving. Squeeze theorem is such an example

**Theorem 2.1 (Squeeze Theorem) **
Suppose that
\[
f(x)\leq g(x)\leq h(x)
\]
and
\[
\lim\limits_{x\to c}f(x)=L=\lim\limits_{x\to c}h(x).
\]
Then
\[
\lim\limits_{x\to c}g(x)=L.
\]

Let’s use Maple to understand the statement.

**Example 2.7 **
Graph the functions \(f(x)=-x\), \(g(x)=x\cos\frac1x\) and \(h(x)=x\) in the same coordinate system. What’s the limit of \(g(x)\) as \(x\) approaches \(0\).

*Solution. *
We use the `plot()`

command to graph the functions together.

`plot([-x, x*cos(1/x), x], x=-1..1, discont, color=[red, blue, red])`

From the graph, we see that the \(\lim\limits_{x\to c}f(x)=0\) as it is squeezed by two limits which are both \(0\).

**Exercise 2.7 **
Graph the functions \(f(x)=-x\), \(g(x)=x\sin\frac1x\) and \(h(x)=x\) in the same coordinate system. What’s the limit of \(g(x)\) as \(x\) approaches \(0\).

## 2.6 Continuity

A function \(f\) is continuous at \(x=a\) if \(f(a)\) is defined and \(\lim\limits_{x\to a}f(x)=f(a)\). Intuitively, a function is continuous if the graph has no hole or jump.

**Example 2.8 **
Use graph to determine if the function
\[
f(x)=
\begin{cases}
x & x\le -1 \\
1/(x-1) & -1<x<1\\
3-x & 1\le x\le 2\\
\sin(x - 2) + 1 & x>2\\
-2 & x=2
\end{cases}
\]
is continuous over \((-\infty, \infty)\). Find the discontinuities and verify them using the definition and properties of continuity.

*Solution. *
First we define the function.

```
f := x -> piecewise(x <= -1, x, -1 < x and x < 1, 1/(x - 1),
1 <= x and x < 2, 3 - x, 2 < x, sin(x - 2) + 1, -2)
```

We first check visually whether the graph has holes or jumps.

`plot(f(x), discont=[showremovable], x=-5..5, smartview=true)`

In the above command, the option `discont = [showremovable]`

is used to show removable discontinuities.

From the graph, we can tell that the function has discontinuities which can also be found using Maple.

`discont(f(x), x)`

Maple gives three discontinuities \(\{-1, 1, 2\}\).

Let’s first find limits at all three values.

```
discontset:=[-1, 1, 2];
for a in discontset do limit(f(x), x=a) end do;
```

You will find that limits do not exist at \(-1\) and \(1\). The limit at \(2\) is \(1\). However, \(g(2)=-2\). So \(f\) has three discontinuities at \(x=-1\), \(x=1\) and \(x=2\).

**Exercise 2.8**Determine if the function \[ f(x)= \begin{cases} -x^2+2 & x\le 1 \\ \frac{1}{x-2}& 1<x<2\\ 2\cos x-1 & \text{otherwise} \end{cases} \] is continuous over \((-\infty, \infty)\). Find all discontinuities if they exist and verify them using the definition.

## 2.7 Continuity and Limit of a Composite Function

Continuous functions behave under composition. The composition of continuous functions is still continuous over its domain. More generally,

**Theorem 2.2**Let \(f\) be a function continuous at \(x=L\). Suppose that \(\lim\limits_{x\to c}g(x)=L\). Then \[ \lim{x\to c}f(g(x))=f(L). \]

In general, the limit does not commute with composition.

**Example 2.9 **
Let
\[
f(x)=x^3\quad \text{and}\quad
g(x)=
\begin{cases}
1 & x=0\\
2x-1 & \text{otherwise}
\end{cases}
\]

Verify that \[ \lim_{x\to 0}f(g(x))=f(\lim_{x\to 0}g(x)). \]

Is is true that \[ \lim_{x\to 0}g(f(x))=g(\lim_{x\to 0}f(x))? \] *Solution. * We first define \(f\) and \(g\) in Maple

```
f:=x->x^2;
g:=x->piecewise(x=0, 1, 2x-1);
```

Evaluate limits for \(f\circ g\).

```
limit(f(g(x)), x=0);
f(limit(g(x), x=0));
```

The results verify that the limit commutes with composition if the outside function is continuous.

Evaluate limits for \(g\circ f\)

```
limit(f(g(x)), x=0);
f(limit(g(x), x=0));
```

The results show that the limit may not commute with composition if the outside function is not continuous.

**Exercise 2.9 **
Find three functions \(f\), \(g\), \(h\) and a value \(c\) such that \(f(g(x))\) is continuous,
\[
\lim\limits_{x\to c}f(h(x))=f(\lim\limits_{x\to c}h(x))
\]
and
\[
\lim\limits_{x\to c}h(g(x))\neq h(\lim\limits_{x\to c}g(x)).
\]

## 2.8 Intermediate Value Theorem

A very important result about continuity is the intermediate value theorem (IVT for short).

**Theorem 2.3 (Intermediate Value Theorem) **Let \(f\) be a function continuous over the interval \([a, b]\). Suppose that \(f(a)\neq f(b)\) and \(N\) is a number between \(f(a)\) and \(f(b)\). Then there exists a number \(c\in (a, b)\) such that \(f(c)=N\).

In particular, if \(f(a)f(b)<0\) then there exists a number \(c\) such that \(f(c)=0\).

**Example 2.10 **
Determine whether the equation
\[
\sin^2x+2x-1=0
\]
has the solution in \((-1, 1)\). Estimate the solution if it exists.

*Solution. *
We first use the left hand side of the equation to define a function \(eql\).

eql:=x->sin(x)^2 + 2*x - 1

Now lets verify that \(eql\) is continuous over \([-1,1]\) using the command `iscont()`

.

`iscont(eql(x), x=-1..1)`

The result is `True`

. We may apply the IVT. Let’s check if the value \(eql(-1)\cdot eql(1)<0\).

`evalf(eql(-1)*eql(1))`

Here `evalf()`

convert the symbolic answer to the (approximate) numerical value.

Since the product is negative, applying the IVT, we know there exists a solution between \((-1, 1)\).

This can also be seen from the graph using the command.

`plot(eql(x), x=-1..1)`

To estimate the solution, you may use the Maple command

`evalf(solve(eql(x) = 0, x))`

or you may repeatedly apply IVT to find an approximate solution.

```
m:=10;
a:=-1;
b:=1;
for n from 1 to m do
c[n] := (a + b)/2;
if evalf(eql(c[n])*eql(a)) < 0 then
b := c[n];
else
a := c[n];
end if;
evalf(eql(c[n]));
end do;
```

**Exercise 2.10 **
Find an integer \(k\) such that the equation
\[
\cos^2x + 3x - 2=0
\]
has a solution in \((k, k+1)\). Estimate the solution.