- Definition of the limit at $+\infty$
-
Definition.
Let $D$ be a subset of $\mathbb R$ and let $f: D \to \mathbb R$ be a function. Let $L$ be a real number. We say that the function $f$
has a limit $L$ at $+\infty$ if the following two conditions are satisfied:
- I.
There exists $X_0 \in \mathbb R$ such that $f(x)$ is defined for every $x \geq X_0.$
- II.
For every $\epsilon \gt 0$ there exists $X(\epsilon) \in \mathbb R $ such that $X(\epsilon) \geq X_0$ and
$x \gt X(\epsilon)$ implies $|f(x) - L | \lt \epsilon$.
The fact that the function $f$ has a limit $L$ at infinity is symbolically expressed as
$\displaystyle\lim_{x\to +\infty} f(x) = L$.
- Narrative
-
-
I mentioned in class that I have seen an engineering paper in which the author states $\tanh(8) = 1.$ But, we proved that for all real numbers $x$ we have $-1 \lt \tanh(x) \lt 1.$ Although, the engineer was formally wrong, his writing was justified if an error of less than $10^{-6}$ was tolerable. To justify this notice that Mathematica gives the approximation for $\tanh(8)$ to $7$ decimal places to be $0.9999998.$ Therefore
\[
0.99999974 \lt \tanh(8) \lt 0.99999985
\]
and consequently
\[
1- \tanh(8) \lt 1- 0.99999985 = 1.5\times 10^{-7} \lt 10^{-6}.
\]
Since the hyperbolic tangent is an increasing function (as proved below) we also have that
\[
1- \tanh(x) \lt 10^{-6}
\]
for all $x \gt 8.$
-
Now we could ask whether a more accurate engineer would also be justified to write $\tanh(x) = 1$ for some $x$? This bring us close to the definition of limit. Roughly speaking the definition of limit tells that no matter how accurate an engineer (this is measured by $\epsilon \gt 0$) there exist real numbers $x$ (this is determined by $X(\epsilon)$ in the definition) for which that engineer could write $\tanh(x) = 1.$
-
Here I prove that $\tanh$ is an increasing function. Let $x_1, x_2$ be real numbers such that $x_1 \lt x_2.$ To prove that $\tanh$ is increasing I will simplify the expression $\tanh(x_2) - \tanh(x_1):$
\begin{align*}
\tanh(x_2) - \tanh(x_1) & = \frac{e^{x_2} - e^{-x_2}}{e^{x_2} + e^{-x_2}} - \frac{e^{x_1} - e^{-x_1}}{e^{x_1} + e^{-x_1}} \\
& = \frac{e^{2x_2} - 1}{e^{2x_2} + 1} - \frac{e^{2x_1} - 1}{e^{2 x_1} + 1} \\
& = \frac{\bigl(e^{2x_2} - 1\bigr)\bigl(e^{2 x_1} + 1\bigr) - \bigl(e^{2x_1} - 1\bigr)\bigl(e^{2 x_2} + 1\bigr)}{\bigl(e^{2x_2} + 1\bigr)\bigl(e^{2 x_1} + 1\bigr)} \\
& = 2 \frac{e^{2x_2} - e^{2x_1}}{\bigl(e^{2x_2} + 1\bigr)\bigl(e^{2 x_1} + 1\bigr)}.
\end{align*}
Since the exponential function is increasing and $2x_2 \gt 2x_1$, the last expression is positive. Hence, $\tanh$ is an increasing function.
Let us calculate the inverse $\tanh^{-1}$ of which will prove that $\tanh$ is a bijection. This function is sometimes denoted arctanh.l
-
Since $\tanh$ is an increasing function it is an injection. To prove that $\tanh$ is a surjection, let $y \in (-1,1)$ and let us solve $\tanh(x) = y.$ That is solve
\[
\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = y.
\]
Multiplying both sides by $e^{x} + e^{-x} \gt 0$ we get
\[
e^{x} - e^{-x} = y \bigl( e^{x} + e^{-x} \bigr).
\]
Multiplying by $e^{-x} \gt 0$ we get
\[
e^{2x} - 1 = y \bigl( e^{2x} + 1 \bigr)
\]
and simplifying yields
\[
e^{2x} \bigl(1-y\bigr) = y + 1.
\]
Separating $x$ and $y$ in the last equality yields
\[
e^{2x} = \frac{1+y}{1-y}.
\]
Since $y \in (-1,1)$ we have $0 \lt 1-y \lt 2$, $0 \lt 1+y \lt 2$ and $\bigl(1+y\bigr)/\bigl(1-y\bigr) \gt 0.$
Hence
\[
x = \frac{1}{2}\, \ln \frac{1+y}{1-y}.
\]
This proves
\[
\tanh^{-1}(x) = \frac{1}{2}\, \ln \frac{1+x}{1-x} \quad \text{for all} \quad x \in (-1,1).
\]
$\tanh$ in navy blue and its inverse $\tanh^{-1}$ in maroon
|
- Example 1
-
Problem. Use the definition of the limit at $+\infty$ to prove that
\[
\lim_{x\to+\infty} \tanh(x) = 1.
\]
Proof. Let $\epsilon \gt 0$ be arbitrary. If $\epsilon \geq 2$, then for all $x \in \mathbb R$ we have
\[
|\tanh(x) - 1| = 1 - \tanh(x) \lt 2 \leq \epsilon.
\]
Therefore, if $\epsilon \geq 2$ we can take $X(\epsilon) = 1.$ Let $\epsilon \in (0,2).$ Then $1-\epsilon \in (-1,1).$ We need to solve
\[
|\tanh(x) - 1| = 1 - \tanh(x) \lt \epsilon.
\]
Simplifying the last inequality is equivalent to
\[
1-\epsilon \lt \tanh(x).
\]
Since $\tanh^{-1}$ is also an increasing function, the last inequality is equivalent to
\[
\tanh^{-1}\bigl(1-\epsilon\bigr) \lt x.
\]
Since
\[
\tanh^{-1}\bigl(1-\epsilon\bigr) = \frac{1}{2}\, \ln \frac{2 - \epsilon}{\epsilon},
\]
we have proved that
\[
x \gt \frac{1}{2}\, \ln \frac{2 - \epsilon}{\epsilon} \qquad \text{implies} \qquad |\tanh(x) - 1| \lt \epsilon.
\]
This shows that for a given $\epsilon \gt 0$ we can take $X(\epsilon) = 1$ if $\epsilon \geq 2$ and $X(\epsilon) = \frac{1}{2}\, \ln \bigl(\bigl(2 - \epsilon\bigr)/\epsilon\bigr)$ if $\epsilon \lt 2.$
Let us test this conclusion on $\epsilon = 10^{-6}.$ Then, approximated to 10 decimal places
\[
X(10^{-6}) = \frac{1}{2}\, \ln \frac{2 - 10^{-6}}{10^{-6}} \approx 7.2543286193.
\]
Hence
\[
x \gt X(10^{-6}) \approx 7.2543286193 \qquad \text{implies} \qquad 1 - \tanh(x) \lt 10^{-6}.
\]
Since $8 \gt X(10^{-6})$ the last implication justifies sloppy engineer's (one who tolerates $10^{-6}$ error) writing $\tanh(8) = 1.$
- Example 2
-
Problem.
Use the definition of the limit at infinity to prove
\[
\lim_{x\to+\infty} \frac{x}{x+\sin(x)} = 1.
\]
Proof. Clearly the function $x/\bigl(x+\sin(x)\bigr)$ is defined for all $x \geq 2.$ So, we can take $X_0 = 2$ in the definition of limit.
Let $\epsilon \gt 0.$ We need to solve
\[
\left| \frac{x}{x+\sin(x)} - 1 \right| \lt \epsilon
\]
to find $X(\epsilon).$ First simplify (here we assume that $x \geq 2$)
\[
\left| \frac{x}{x+\sin(x)} - 1 \right| = \left| \frac{x-x-\sin(x)}{x+\sin(x)} \right| = \frac{|-\sin(x)|}{x+\sin(x)}
\]
Still, solving
\[
\frac{|-\sin(x)|}{x+\sin(x)}\lt \epsilon
\]
is not possible. Therefore we replace $\frac{|-\sin(x)|}{x+\sin(x)}$ with something slightly bigger and solvable:
\[
\frac{|-\sin(x)|}{x+\sin(x)} \leq \frac{1}{x-1} \lt \epsilon
\]
Now, solving
\[
\frac{1}{x-1} \lt \epsilon
\]
for $x \geq 2$ will lead to $X(\epsilon)$: Multiply be $x-1 \gt 0$ and divide by $\epsilon \gt 0$ to get
\[
\frac{1}{\epsilon} \lt x - 1.
\]
Thus, the solution is
\[
x \gt \frac{1}{\epsilon} + 1.
\]
Since we need $X(\epsilon) \geq 2$, we will define
\[
X(\epsilon) = \max\left\{\frac{1}{\epsilon} + 1,2 \right\}.
\]
It remains to prove
$x \gt X(\epsilon)$ implies $\left| \frac{x}{x+\sin(x)} - 1 \right| \lt \epsilon.$
I hope that you can provide this proof. An important fact that you will use in this proof is the inequality that we established above: for all $x \geq 2$ the following inequality holds:
\[
\frac{|-\sin(x)|}{x+\sin(x)} \leq \frac{1}{x-1}.
\]
Limit at a real number $c$
- Definition of the limit at a real number $c$
-
Definition.
Let $D \subseteq \mathbb R$, let $c, L \in \mathbb R$ and let $f: D \to \mathbb R$ be a function. We say that the function $f$
has a limit $L$ at $c$ if the following two conditions are satisfied:
- I.
There exists $\delta_0 \gt 0$ such that $f(x)$ is defined for every $x \in (c-\delta_0,c)\cup(c,c+\delta_0)$; in other words $(c-\delta_0,c)\cup(c,c+\delta_0) \subseteq D.$
- II.
For every $\epsilon \gt 0$ there exists $\delta(\epsilon)$ such that $0 \lt \delta(\epsilon) \leq \delta_0$ and
$0 \lt |x-c| \lt \delta(\epsilon)$ implies $|f(x) - L | \lt \epsilon.$
The fact that the function $f$ has a limit $L$ at $a$ is symbolically expressed as
$\displaystyle\lim_{x\to a} f(x) = L.$
- Example 1
-
Problem. Use the definition of limit to prove that $\displaystyle \lim_{x\to 2} x^2 = 4$.
How does this fit into the definition of limit? Here $D = \mathbb R$, $f(x) = x^2$, $c=2$ and $L = 4.$
Proof.
To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $f(x)$ is defined for all $x \in (1,2)\cup(2,3).$
Let $\epsilon \gt 0$ be arbitrary. Our next objective is to discover $\delta(\epsilon).$ We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x-2| \lt \delta(\epsilon)$ imply $|x^2 - 4 | \lt \epsilon.$
To discover $\delta(\epsilon)$ we need the following fact.
Fact A.
If $|x-2| \lt 1$, then $|x^2 - 4| \leq 5 |x - 2|.$
To toggle the proof of Fact A click
Assume that $|x-2| \lt 1.$ Then $1 \lt x \lt 3.$ Consequently, $3 \lt x+2 \lt 5.$ Since $x+2 \gt 0,$ we have $|x+2| = x+2$ and hence $|x+2| \lt 5.$ Next recall the difference of two squares formula and rules for the absolute value function to simplify
\[
| x^2 - 4 | = |(x+2)(x-2)| = |x+2|\, |x-2|.
\]
Since $|x+2| \lt 5$ the last identity yields $|x^2 - 4| \leq 5 |x - 2|.$ This proves Fact A.
Our goal is to solve have $| x^2 - 4 | \lt \epsilon$ for $|x-2|.$ This is somewhat complicated. However,
Fact A yields that the solutions of $5 |x-2| \lt \epsilon$ will also be solutions of $| x^2 - 4 | \lt \epsilon.$ Solving $5 |x-2| \lt \epsilon$ for $|x-2|$ is easy. The solution is $|x-2| \lt \epsilon/5.$ Therefore we can choose
\[
\delta(\epsilon) = \min\{\epsilon/5,1\} = \begin{cases} 1 & \text{if} \ \ \epsilon \geq 5 \\
\epsilon/5 & \text{if} \ \ \epsilon \lt 5 \end{cases}.
\]
This $\delta(\epsilon)$ has the following properties:
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x-2| \lt \delta(\epsilon)$ implies $|x^2 - 4 | \lt \epsilon.$
Why? Since both $\epsilon/5$ and $1$ are positive we have $\delta(\epsilon) \gt 0.$ Since $\min\{\epsilon/5,1\} \leq 1$ we have $\delta(\epsilon) \leq 1.$ Thus $0 \lt \delta(\epsilon) \leq 1.$
To prove the implication
\[
0 \lt |x-2| \lt \delta(\epsilon) \quad \Rightarrow \quad |x^2 - 4 | \lt \epsilon
\]
assume that $0 \lt |x-2| \lt \min\{\epsilon/5,1\}.$ Then $|x-2| \lt 1.$ By Fact A this implies $|x^2 - 4| \leq 5 |x - 2|.$ The assumption $|x-2| \lt \min\{\epsilon/5,1\}$ also implies $|x-2| \lt \epsilon/5$ and therefore $5 |x-2| \lt \epsilon.$ Since we already proved $|x^2 - 4| \leq 5 |x - 2|$, we thus have $|x^2 - 4 | \lt \epsilon.$
- Example 2
-
Problem. Use the definition of limit to prove that $\displaystyle \lim_{x\to 0} \cos x = 1.$
How does this fit into the definition of limit? Here $D = \mathbb R$, $f(x) = \cos(x)$, $c=0$ and $L = 1.$
Proof.
To prove the given limit, in the definition of limit we can take $\delta_0 = 1 \lt \pi/3,$ since clearly $\cos(x)$ is defined for all $x \in (-1,0)\cup(0,1).$
Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos(x) - 1 | \lt \epsilon.$
To discover $\delta(\epsilon)$ we need the following fact.
Fact B.
If $|x| \lt 1$, then $|\cos(x) - 1| \leq |x|.$
To toggle the proof of Fact B click
First assume that $0 \lt x \lt 1.$ Then we can relay on the figure below. This figure shows a part of the unit circle. The number $x$ is represented by the teal circular arc length between points $A$ and $C.$ Then, by the definition of the cosine function the blue length of the line segment $OB$ equals $\cos x.$ The triangle $ABC$ is a right triangle whose one leg is the line segment $BC$ and whose hypotenuse is $AC.$ Therefore the length $BC$, which equals $1-\cos x$, is less than the length of the line segment $AC.$ Further the length of the line segment $AC$ is less than arc length of the teal circular arc between $A$ and $C.$ Since the arc length of the teal circular arc equals $x$, this proves $1 - \cos x \lt x.$ Similarly, if $-1 \lt x \lt 0$, we have $1 - \cos x \lt -x.$ Thus, in either case $1 - \cos x \lt |x|.$ Since $\cos(0) =1$ and $|\cos x - 1| = 1-\cos x$ this proves
Fact B.
Figure for $\displaystyle \lim_{x\to 0} \cos(x) = 1$
|
Our goal is to solve have $| \cos(x) - 1 | \lt \epsilon$ for $|x|.$ This is somewhat complicated. However,
Fact B yields that the solutions of $|x| \lt \epsilon$ will also be solutions of $| \cos(x) - 1 | \lt \epsilon.$ But $|x| \lt \epsilon$ is already solved for $|x|.$ Therefore
Fact B yields that $|x| \lt \epsilon$ implies $| \cos(x) - 1 | \lt \epsilon.$ But, to use
Fact B we need $|x| \lt 1.$ Therefore we set
\[
\delta(\epsilon) = \min\{\epsilon,1\}.
\]
This $\delta(\epsilon)$ has the following properties:
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $|\cos(x) - 1 | \lt \epsilon.$
Why? Since both $\epsilon$ and $1$ are positive we have $\delta(\epsilon) \gt 0.$ Since $\min\{\epsilon,1\} \leq 1$ we have $\delta(\epsilon) \leq 1.$ Thus $0 \lt \delta(\epsilon) \leq 1.$
To prove the implication
\[
0 \lt |x| \lt \delta(\epsilon) \quad \Rightarrow \quad |\cos(x) - 1 | \lt \epsilon
\]
assume that $0 \lt |x| \lt \min\{\epsilon,1\}.$ Then $|x| \lt 1.$ By Fact B this implies $|\cos(x) - 1| \leq |x|.$ The assumption $|x| \lt \min\{\epsilon,1\}$ also implies $|x| \lt \epsilon.$ Since by Fact B $|\cos(x) - 1| \leq |x|$, we thus have $|\cos(x) - 1| \lt \epsilon.$
- Example 3
-
Problem. Use the definition of limit to prove that $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1.$
How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{\sin x}{x}$, $c=0$ and $L = 1.$
Proof. To prove the given limit, in the definition of limit we can take $\delta_0 = 1 \lt \pi/3,$ since clearly $\displaystyle \frac{\sin x}{x}$ is defined for all $x \in (-1,0)\cup(0,1).$
Our next objective is to discover $\delta(\epsilon).$ Let $\epsilon \gt 0$ be arbitrary. We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{\sin x}{x} - 1 \right| \lt \epsilon.$
To discover $\delta(\epsilon)$ we need the following fact.
Fact C.
If $0 \lt |x| \lt 1$, then $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt |x|.$
To toggle the proof of Fact C click
The proof of
Fact C is slightly more complicated than the proof of
Fact B.
We first prove
Fact C-1.
If $0 \lt x \lt 1$, then $\displaystyle \cos(x) \lt \frac{\sin x}{x} \lt 1.$
To prove
Fact C-1 assume that $0 \lt x \lt 1$ and rely on the figure below. This figure shows a part of the unit circle. The number $x$ is represented by the teal circular arc length between points $A$ and $C.$ By the definition of the cosine function the length of the navy blue line segment $OB$ equals $\cos x$ and by the definition of the sine function the green length of the line segment $AB$ equals $\sin x.$ Now compare the areas of the triangle $OCA$, the yellow area of the
circular sector identified by points $OCA$ and the area of the triangle $OCD.$ Clearly these three areas are listed in the increasing order. By calculating each of the three areas we conclude
\[
\frac{1}{2} \sin x \lt \frac{1}{2} x \lt \frac{1}{2} \tan x.
\]
Dividing the preceding inequalities by $\dfrac{1}{2} \sin x \gt 0$ and taking reciprocals yields
\[
\cos x \lt \frac{\sin x}{x} \lt 1.
\]
This proves
Fact C-1.
To prove Fact C assume again that $0 \lt x \lt 1.$ By Fact B we have that $1- \cos(x) \lt |x|,$ that is $1- |x| \lt \cos(x).$ Together with Fact C-1 this yields \[
1-|x| \lt \frac{\sin x}{x} \lt 1.
\]
Since all the functions in the preceding inequality are even, the inequality also holds for negative numbers $x \in (-1,0).$ Now subtracting $1$ in each term of $1-|x| \lt (\sin x)/x \lt 1$ yields $-|x| \lt (\sin x)/x -1 \lt 0,$ and multiplying by $-1$ finally gives $0 \lt 1- (\sin x)/x \lt |x|.$ This proves Fact C.
Figure for $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = 1$
|
Our goal is to solve have $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon$ for $|x|.$ For this we can use
Fact C and repeat the reasoning from
Example 2 to conclude that in this case we can also set
\[
\delta(\epsilon) = \min\{\epsilon,1\}.
\]
This $\delta(\epsilon)$ has the following properties:
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{\sin x}{x} -1 \biggr| \lt \epsilon.$
Why? The reasoning is identical to the reasoning presented in Example 2.
- Example 4
-
Problem. Use the definition of limit to prove that $\displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x^2} = \frac{1}{2}.$
How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{1-\cos(x)}{x^2}$, $c=0$ and $L = 1/2.$
Proof. To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $\displaystyle \frac{1-\cos(x)}{x^2}$ is defined for all $x \in (-1,0)\cup(0,1).$
Let $\epsilon \gt 0$ be arbitrary. Our next objective is to discover $\delta(\epsilon).$ We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{1-\cos(x)}{x^2} - \frac{1}{2} \right| \lt \epsilon.$
To discover $\delta(\epsilon)$ we need the following fact.
Fact D.
If $0 \lt |x| \lt 1$, then $\displaystyle \biggl|\frac{1-\cos(x)}{x^2} - \frac{1}{2} \biggr| \lt |x|.$
To toggle the proof of Fact D click
The proof of
Fact D is slightly more complicated than the proof of
Fact C.
As in the proofs of
Fact B and
Fact C we will use a geometric argument deduced from the unit circle.
In the figure below we consider $x \in (0,1).$ Further we consider the lengths of the line segments $\overline{AB}$, $\overline{AC}$ and the length of the circular arc from $A$ to $C.$ We see that these three lengths are listed in increasing order and they calculate to be:
\[ \tag{#}
\sin x \lt \sqrt{2 - 2 \cos x} \lt x.
\]
Here, the length of $\overline{AC}$ is calculated by the Pythagorean theorem
\[
\sqrt{(\sin x)^2 +(1- \cos x)^2} = \sqrt{(\sin x)^2 +1- 2 \cos x +(\cos x)^2} = \sqrt{2 - 2 \cos x}
\]
|
Figure for $\displaystyle \lim_{x\to 0} \frac{1- \cos x}{x^2} = \frac{1}{2}$
|
Squaring the inequalities in (#) we get
\[
(\sin x)^2 \lt 2 - 2 \cos x \lt x^2.
\]
Now, dividing the preceding inequalities by the positive number $2 x^2$ we get
\[ \tag{*}
\frac{1}{2} \left(\frac{\sin x}{x}\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}.
\]
In the proof of
Fact C we proved that
\[
1 - |x| \lt \frac{\sin x}{x}.
\]
Since $x \in (0,1)$ both quantities in the preceding inequality are positive. Therefore we can square both sides of the inequality and divide by $2$ to get
\[
\frac{1}{2} \left(1-|x|\right)^2 \lt \frac{1}{2} \left(\frac{\sin x}{x}\right)^2.
\]
Together with inequalities (*), the preceding inequality yields
\[
\frac{1}{2} \left(1-|x|\right)^2 \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}.
\]
Since
\[
\frac{1}{2} - |x| \lt \frac{1}{2} \left(1-|x|\right)^2,
\]
we conclude
\[
\frac{1}{2} - |x| \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}.
\]
In our reasoning so far we assumed that $x \in (0,1).$ But, since all the functions in the preceding expression are even, the inequalities also hold for $x \in (-1,0)$ as well. That is, for all $x \in (-1,0) \cup (0,1)$ we have
\[
\frac{1}{2} - |x| \lt \frac{1 - \cos x}{x^2} \lt \frac{1}{2}.
\]
Using the last two inequalities, for all $x \in (-1,0) \cup (0,1)$ we deduce
\begin{align*}
\left| \frac{1-\cos(x)}{x^2} - \frac{1}{2} \right| & = \frac{1}{2} - \frac{1 - \cos x}{x^2} \\
& \lt \frac{1}{2} - \left( \frac{1}{2} - |x| \right) \\
& = |x|.
\end{align*}
This proves
Fact D.
Our goal is to solve have $\displaystyle \biggl| \frac{1- \cos(x)}{x^2} -\frac{1}{2} \biggr| \lt \epsilon$ for $|x|.$ For this we can use
Fact D and repeat the reasoning from
Example 2 to conclude that in this case we can also set
\[
\delta(\epsilon) = \min\{\epsilon,1\}.
\]
This $\delta(\epsilon)$ has the following properties:
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{1- \cos(x)}{x^2} - \frac{1}{2} \biggr| \lt \epsilon.$
Why? The reasoning is identical to the reasoning presented in Example 2.
- Example 5
-
Problem. Use the definition of limit to prove that $\displaystyle \lim_{x\to 0} \frac{1-\cos(x)}{x} = 0.$
How does this fit into the definition of limit? Here $D = {\mathbb R}\setminus\{0\}$, $\displaystyle f(x) = \frac{1-\cos(x)}{x}$, $c=0$ and $L = 0.$
Proof. To prove the given limit, in the definition of limit we can take $\delta_0 = 1,$ since clearly $\displaystyle \frac{1-\cos(x)}{x}$ is defined for all $x \in (-1,0)\cup(0,1).$
Let $\epsilon \gt 0$ be arbitrary. Our next objective is to discover $\delta(\epsilon).$ We need to discover $\delta(\epsilon)$ such that
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \left| \frac{1-\cos(x)}{x} - 0 \right| \lt \epsilon.$
To discover $\delta(\epsilon)$ we need the following fact.
Fact E.
If $0 \lt |x| \lt 1$, then $\displaystyle \biggl|\frac{1-\cos(x)}{x} \biggr| \lt \frac{1}{2} |x|.$
To toggle the proof of Fact E click
The proof of Fact E uses an inequality that was established in the proof of Fact D. Let $0 \lt |x| \lt 1$, that is let $x \in (-1,0)\cup(0,1).$ In the proof of Fact D we proved that
\[
\frac{1-\cos(x)}{x^2} \lt \frac{1}{2}.
\]
Clearly
\[
0 \lt \frac{1-\cos(x)}{x^2},
\]
we have
\[
0 \lt \frac{1-\cos(x)}{x^2} \lt \frac{1}{2}.
\]
Multiplying the preceding expression by $|x| \gt 0$ we get
\[
0 \lt \frac{1-\cos(x)}{|x|} \lt \frac{1}{2}|x|.
\]
Since
\[
\biggl|\frac{1-\cos(x)}{x} \biggr| = \frac{1-\cos(x)}{|x|},
\]
this proves Fact E.
Our goal is to solve have $\displaystyle \biggl| \frac{1- \cos(x)}{x} -0 \biggr| \lt \epsilon$ for $|x|.$ For this we can use
Fact E and discover that we need to set
\[
\delta(\epsilon) = \min\{2\epsilon,1\}.
\]
This $\delta(\epsilon)$ has the following properties:
$0 \lt \delta(\epsilon) \leq 1$ and
$0 \lt |x| \lt \delta(\epsilon)$ implies $\displaystyle \biggl|\frac{1- \cos(x)}{x} - 0 \biggr| \lt \epsilon.$
Why? The reasoning is quite similar to the reasoning presented in Example 2. You just need to use Fact E instead of Fact B.
- Lipschitz condition for the existence of limit
-
All the above examples fit one pattern which is summarized in the following condition.
Lipschitz condition for the existence of limit.
Let $c \in \mathbb R$ and $D \subset \mathbb R$. Let $f : D \to \mathbb R$ be a function. We say that $f$ satisfies Lipschitz condition at $c$ if there exist $\delta_0 \gt 0$, $K \gt 0$ and $L \in \mathbb R$ such that $(c-\delta_0,c)\cup (c,c+\delta_0) \subseteq D$ and for every $x \in (c-\delta_0,c)\cup (c,c+\delta_0)$ we have
\[
\bigl| f(x) - L \bigr| \leq K\, |x-c|.
\]
If you look carefully through our examples you will notice that in each example we in fact proved that the function that was considered satisfies the Lipschitz condition for the existence of limit. In our examples we the established inequality to prove the existence of limit using the definition. However, the following general theorem holds.
Theorem.
Let $c \in \mathbb R$ and $D \subset \mathbb R$. Let $f : D \to \mathbb R$ be a function. Assume that $f$ satisfies Lipschitz condition at $c$ with constants $\delta_0 \gt 0$, $K \gt 0$ and $L \in \mathbb R$. Then
\[
\lim_{x\to c}f(x) = L.
\]
I hope that you can prove this theorem on your own.
- From the textbook
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In the Calculus textbook read Sections 1.7 and 1.8. Do the problems: 1.7 19, 20, 21, 22, 23, 24, 30, 1.8 1, 7, 8, 10, 14, 20, 46, 47, 57.