Limit of Functions

Intuitive approach to limit As $x$ approaches $x_{0}$, $f(x)$ approaches $l$, Then we say $l$ is the limit of $f(x)$ as $x$ approaches $x_{0}$ and denote

$$\lim_{x \rightarrow x_{0}} f(x) = l$$

Expression of limit $\lim_{x \rightarrow x_{0}} f(x) = l$ or $x \to x_{0} \longrightarrow f(x) \to l$.

NOTE $x$ approaches $x_{0}$ is the same as $\vert x - x_{0}\vert$ approaches 0. Similarly, $f(x)$ approaches $l$ is the same as $\vert f(x) - l\vert$ approaches 0.

Understanding of limit The concept of infinitesimal can be explained by saying that every small number you choose, we can choose smaller number.

Limit properties

Theorem 1..6   Let $${\lim_{x \rightarrow x_{0}}f(x) = l, \lim_{x \rightarrow x_{0}}g(x) = m}$$ and $c$ be constan, Then we have the followings:

$${1. \lim_{x \rightarrow x_{0}}(f(x) \pm g(x)) = l \pm m}$$

$${2. \lim_{x \rightarrow x_{0}}cf(x) = cl}$$

$${3. \lim_{x \rightarrow x_{0}}(f(x)g(x)) = lm}$$

$${4. \lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)} = \frac{l}{m} {\rm provided} m \neq 0}$$

NOTE Limit of functions obey the four rules of arithmetic provided the denominator is not 0.

Example 1..21   Find the limit of the following.
$${1. \lim_{x \to 2} \frac{x^2 - 3x + 2}{x^2 - 4}}$$ $${2. \lim_{x \to 0} \frac{\sqrt{x + 4} -2}{x}}$$

Note that $x \to 2 \longrightarrow x^2 - 3x +2 \to 0$ and $x^2 - 4 \to 0$. In other words, both the denominator and the numerator have the common factor $(x-2)$.

SOLUTION1.

$$\lim_{x \to 2}\frac{x^2 - 3x + 2}{x^2 - 4} = \lim_{x \to 2}\frac{... ...-2)} = \lim_{x \to 2}\frac{x-1}{x + 2} = \frac{1}{4}\ensuremath{ \blacksquare}$$

As $x \to 0$, we have $\sqrt{x+4} - 2 \to 0$ and $x \to 0$. Thus we can factorby $x$.

2.

$$\lim_{x \to 0} \frac{\sqrt{x + 4} -2}{x} = \lim_{x \to 0} \frac{(\sqrt{x + 4} -2)(\sqrt{x+4} + 2)}{x(\sqrt{x+4} + 2)}$$

$$= \lim_{x \to 0}\frac{x+4-4}{x(\sqrt{x+4} + 2)} = \lim_{x \to 0}\frac{1}{\sqrt{x+4} + 2} = \frac{1}{4}\ensuremath{ \blacksquare}$$

Exercise 1..21   Find the limit of the following.

$$\lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1 - x}}{x}$$

The function $\sqrt{1+x} - \sqrt{1-x}$ can be rationalize by multiplying $\sqrt{1+x} + \sqrt{1-x}$ to the both numerator and denominator.

SOLUTION

$$\lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1 - x}}{x}$$

$$= \lim_{x \rightarrow 0} \frac{(\sqrt{1+x} - \sqrt{1 - x})(\sqrt{1+x} + \sqrt{1 - x})}{x(\sqrt{1+x} + \sqrt{1 - x})}$$

$$= \lim_{x \rightarrow 0} \frac{{1+x} - (1 - x)}{x(\sqrt{1+x} + \sqrt{1 - x})} = \lim_{x \rightarrow 0} \frac{2x}{x(\sqrt{1+x} + \sqrt{1 - x})}$$

$$= \lim_{x \rightarrow 0} \frac{2}{(\sqrt{1+x} + \sqrt{1 - x})} = 1 \ensuremath{ \blacksquare}$$

Diverges to infinity We write $\lim_{x \rightarrow x_{0}}f(x) = +\infty$ when $f(x)$ gets larger without bound as $x$ approaches $x_{0}$. We write $\lim_{x \rightarrow x_{0}}f(x) = -\infty$ when the value of $f(x)$ is negative and the absolute value gets larger without bound as $x$ approaches $x_{0}$.

Difference of two squares $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$.

NOTE $f(x)$ gets larger without bound means that given any large number $M$, there exists number $N$ such that $f(x) > M$ as $x$ gets larger than $N$ . We write $\lim_{x \rightarrow \infty} f(x) = l$ when $f(x)$ approaches $l$ as $x$ gets larger without bound.

Example 1..22   Find the limit of the following.
$${1. \lim_{x \to \infty}(\sqrt{x+1} - \sqrt{x})}$$ $${2. \lim_{x \to \infty}(\sqrt{1+x+x^2} - \sqrt{1 - x + x^2})}$$

SOLUTION 1. $\lim_{x \to \infty}\sqrt{x+1} = \infty$, $\lim_{x \to \infty}\sqrt{x} = \infty$. Thus it is indeterminate form of $\infty - \infty$.

$$\lim_{x \to \infty}(\sqrt{x+1} - \sqrt{x}) = \lim_{x \to \infty}\frac{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})}{\sqrt{x+1} + \sqrt{x}}$$

$$= \lim_{x \to \infty}\frac{x+1-x}{\sqrt{x+1} + \sqrt{x}} = \lim_{x \to \infty}\frac{1}{\sqrt{x+1} + \sqrt{x}} = 0\ensuremath{ \blacksquare}$$

$\infty - \infty$ The indeterminate form of $\infty - \infty$ has to be rewrite in the indeterminate form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

2. $\lim_{x \to \infty}\sqrt{1+x+x^2} = \infty$, $\lim_{x \to \infty}\sqrt{1-x+x^2} = \infty$. Thus it is indeterminate form of $\infty - \infty$. Now rationalize the fraction $\sqrt{1+x+x^2} + \sqrt{1 - x + x^2}$.

$$\lim_{x \to \infty}(\sqrt{1+x+x^2} - \sqrt{1 - x + x^2})$$

$$= \lim_{x \to \infty}\frac{(\sqrt{1+x+x^2} - \sqrt{1 - x + x^2})(\sqrt{1+x+x^2} + \sqrt{1-x+x^2})}{\sqrt{1+x+x^2} + \sqrt{1-x+x^2}}$$

$$= \lim_{x \to \infty}\frac{1+x+x^2 - (1 - x + x^2)}{\sqrt{1+x+x^2} + \sqrt{1-x+x^2}}$$

$$= \lim_{x \to \infty}\frac{2x}{\sqrt{1+x+x^2} + \sqrt{1-x+x^2}}$$

$$= \lim_{x \to \infty}\frac{2}{\sqrt{\frac{1}{x^2}+\frac{1}{x}+1} + \sqrt{\frac{1}{x^2}-\frac{1}{x}+1}} = 1\ensuremath{ \blacksquare}$$

$\frac{\infty}{\infty}$

$\lim_{x \to \infty}\frac{2x}{\sqrt{1+x+x^2} + \sqrt{1-x+x^2}}$ is in the indeterminate form of $\frac{\infty}{\infty}$. Then factor the fraction by dividing the largest power of $x$.

Exercise 1..22   Find $\lim_{x \to -\infty}(\sqrt{x^2 + 2x + 1} + x + 1)$

Handling $x \to -\infty$ If $x \to -\infty$, then we write $x = -t$ and solve the question.

SOLUTION Put $x = -t$. Then

$$\lim_{x \to -\infty}(\sqrt{x^2 + 2x + 1} + x + 1) = \lim_{t \to \infty}(\sqrt{t^2 - 2t + 1} - (t - 1))$$

$$= \lim_{t \to \infty}(t-1 - t + 1) = 0 \ensuremath{ \blacksquare}$$

Exercise1.22 $\sqrt{x^2 + 2x + 1} = \sqrt{(x+1)^2}$. Note that $x + 1$ is negative. Thus $\sqrt{(x+1)^2} = -(x+1)$.

To find the limit of function, the above theorem is not enogh. For example $${\lim_{x \rightarrow 0}\frac{\sin{x}}{x}}$$ can not be found..

Squeezing theorem

Theorem 1..7   If $f(x) < h(x) < g(x)$ is satisfied for the $\delta$ neighborhood $(x_{0} - \delta, x_{0} + \delta)$ of $x_0$ and

$$\lim_{x \rightarrow x_{0}}f(x) = \lim_{x \rightarrow x_{0}}g(x) = l.$$

Then $${\lim_{x \rightarrow x_{0}}h(x) = l}$$.

NOTE Since $f(x) < h(x) < g(x)$, $f(x) - l < h(x) - l < g(x) - l$. Note that $\vert f(x) - l\vert$ and $\vert g(x) - l\vert$ can be made as small as possible. Thus we can make $\vert h(x) - l\vert$ as small as possible .

Example 1..23   Find $${\lim_{x \rightarrow 0}\frac{\sin{x}}{x}}$$.

Figure 1.39: Example1-23

SOLUTION Take points ${\rm A}(1,0),{\rm P}$ on the unit circle. Now find the intersection of the extended line OP and the line perpendicular to the line OA. We name the intersection B. Also, start from P, find the intersection of the line perpendicular to OA, we name this C. Now we compare the size of the triangles. Then $\triangle {\rm OPC} < {\rm sector OAP} < \triangle {\rm OAB}$. Now for $${0 < x < \frac{\pi}{2}}$$, we have

$$\underbrace{\frac{\cos{x}\cdot\sin{x}}{2}}_{\triangle \mbox{OPC}}... ...derbrace{\frac{\tan{x}}{2} }_{\triangle \mbox{OAB}} = \frac{\sin{x}}{2\cos{x}}.$$

From the first inequality, $${\frac{\sin{x}}{x} < \frac{1}{\cos{x}}}$$, and the second inequality $${\cos{x} < \frac{\sin{x}}{x}}$$. Thus

$$\cos{x} < \frac{\sin{x}}{x} < \frac{1}{\cos{x}}.$$

This inequality holds for $${-\frac{\pi}{2} < x < 0}$$. Now we have $${\lim_{x \to 0}\cos{x} = 1}$$ and $${\lim_{x \to 0}\frac{1}{\cos{x}}} = 1$$. Therefore,

$$\lim_{x \to 0}\frac{\sin{x}}{x} = 1 \ensuremath{ \blacksquare}$$

Area of sector We compare the area of sector with the radius $r$ and the angle $\theta$ with the area of circle with the radius $r$. Then the arc length of the sector is $\theta$. Thus the area of sector is $\pi r^2 \times \frac{\theta}{2\pi r} = \frac{r \theta}{2}$.

Exercise 1..23   Find $${\lim_{x \rightarrow 0}\frac{\sin{2x}}{x}}$$.

SOLUTION $${\lim_{x \rightarrow 0}\frac{\sin{x}}{x}} = 1$$ imples that for $x$ small, $\sin{x}$ and $x$ is about the same.

$$\lim_{x \rightarrow 0}\frac{\sin{2x}}{x} = \lim_{x \rightarrow 0}\frac{\sin{2x}}{2x}\cdot \frac{2x}{x}$$

$$= \lim_{X \rightarrow 0}\frac{\sin{X}}{X}\cdot 2 = 2 \ensuremath{ \blacksquare}$$

Set $X = 2x$. Then $\lim_{X \to 0}\frac{\sin{X}}{X} = 1$.

Example 1..24   Find the limit of the followings. $${1. \lim_{x \rightarrow 0}\frac{\sin{3x}}{\sin{2x}}}$$ $${2. \lim_{x \rightarrow 0}\frac{\tan{3x}}{x}}$$

SOLUTION 1. $${\lim_{x \rightarrow 0}\frac{\sin{3x}}{\sin{2x}} = \lim_{x \righta... ...{3x} \frac{2x}{{\sin{2x}}}\frac{3}{2} = \frac{3}{2}\ensuremath{ \blacksquare}}$$

$\lim_{x \to 0}\frac{\sin{3x}}{\sin{2x}} = \lim_{x \to 0}\frac{\frac{\sin{3x}}{3x}}{\frac{\sin{2x}}{2x}}\cdot \frac{3x}{2x}$.

2.

$$\lim_{x \rightarrow 0}\frac{\tan{3x}}{x} = \lim_{x \rightarrow 0}\frac{\sin{3x}}{x\cos{3x}} = \lim_{x \rightarrow 0}\frac{3\sin{3x}}{3x\cos{3x}} = 3\ensuremath{ \blacksquare}$$

Express $\tan{3x}$ using $\sin$ and $\cos$. Thne $\tan{3x} = \frac{\sin{3x}}{\cos{3x}}$
$\lim_{x \to 0}\cos{3x} = 1$.

Exercise 1..24   Find $${\lim_{x \rightarrow \infty}x\sin{\frac{1}{x}}}$$.

$\infty \cdot 0$ If you have the indeterminate form $\infty \cdot 0$, then express the one of the expression by fraction. Then we can transform into $\frac{0}{0}$ or $\frac{\infty}{\infty}$. SOLUTION As $x \to \infty$, $x\sin{\frac{1}{x}}$ has the indeterminate form $(\infty \cdot 0)$. Then we make this into the indeterminate form of $(\frac{0}{0})$.

$$\lim_{x \rightarrow \infty}x\sin{\frac{1}{x}} = \lim_{x \rightarrow \infty}\frac{\sin{\frac{1}{x}}}{\frac{1}{x}}.$$

Put $\frac{1}{x} = t$. Then as $x \to \infty$, we have $t \to 0$. Thus,

$$\lim_{x \rightarrow \infty}\frac{\sin{\frac{1}{x}}}{\frac{1}{x}} = \lim_{t \to 0}\frac{\sin{t}}{t} = 1\ensuremath{ \blacksquare}$$