Precise Definition of Limits (from English to Mathematics)
Let’s commence by revisiting the intuitive concept of limits. We define the limit of a function as $x$ approaches $a$ to be equal to $L$ if we can make the values of $f(x)$ arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ but not equal to $a$. In other words,
for any given margin of error $\varepsilon > 0$, we can ensure that $|f(x) - L| < \varepsilon$ by restricting $x$ to be sufficiently close to $a$.
| Now, let’s delve into the concept of “sufficiency” within this context. In formal logic, we designate the statement $p$ in the implication ($p \Rightarrow q$) as a “sufficient condition.” In our case, this implies that being “sufficiently close to $a$” can be expressed as the implication ($x$ is close to $a$ $\Rightarrow$ $ | f(x) - L | < \varepsilon$). When we mention “$x$ is close to $a$,” we are essentially saying that “$x$ is near $a$” which, in turn, implies that “$x$ is within a neighborhood of $a$.” |
$x$ is within a neighborhood of $a$ $\Rightarrow$ $|f(x)-L|<\varepsilon$.
Furthermore, we can directly translate the notion of “$x$ is within a neighborhood of $a$” to mean “$x$ is within an open interval containing $a$” with $a$ serving as the center of this open interval. Why? If $a$ resides in an open interval $(c_1, c_2)$, we can determine $\delta = \min(a - c_1, c_2 - a)$, which allows us to substitute the original open interval with $(a - \delta, a + \delta)$. Consequently, after a series of clarifications, the initial statement is equivalent to saying
if $x$ is in $(a - \delta, a + \delta)$, then $|f(x) - L| < \varepsilon.$
This is how we get the precise definition of limits.
| Definition. $\displaystyle\lim_{x\to a}f(x) = L$ exists if for any $\varepsilon>0$, there exists $\delta$ such that $ | f(x)-L | <\varepsilon$ while $ | x-a | <\delta$ ($ | x-a | <\delta$ $\Rightarrow$ $ | f(x)-L | <\varepsilon$). |
Let’s utilize this definition to illustrate why it’s impossible for $f(x) = 0$ to occur infinitely close to $a$ when $\lim_{x\to a} f(x) = L > 0$.
| According to the definition, since the limit exists, we have the statement: “for any $\varepsilon > 0$, there exists $\delta > 0$ …blablabla.” In particular, this statement holds true if we choose $\varepsilon = L/2$. Consequently, there exists some $\delta_0 > 0$ such that for all $x$ satisfying $ | x - a | < \delta_0$, we have |
| $$ | f(x)-L | <\frac{L}{2}.$$ |
| It follows $ | f(x) | >L-\frac{L}{2}>0$. |
Limits Law
Theorem.
Suppose $c$ is a real constant, and both \(\lim_{x\to a}f(x)\text{ and }\lim_{x\to a}g(x)\) exist. Then,
- $\displaystyle\lim_{x\to a} f(x)\pm g(x) = \lim_{x\to a}$
- $\displaystyle\lim_{x\to a} f(x)g(x) = \lim_{x\to a}f(x)\lim_{x\to a}g(x)$
- $\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ if $\displaystyle\lim_{x\to a}g(x)\neq 0$
- $\displaystyle\lim_{x\to a} (f(x))^{(b/c)} = \left(\lim_{x\to a}f(x)\right)^{(b/c)}$ if $c$ is not odd and $b/c$ is in its lowest terms (最簡分數形式).
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If absolute value is involved in $f(x)$, for example, in expressions like $(x^2-1)/ x-1 $, then it is advisable to use conditional functions and left-right limits to evaluate its limit. - The squeeze theorem is frequently useful when we encounter limits of functions that involve a function tending to zero multiplied by another function with bounded values, e.g. $x\sin(1/x)$, $x\cos(1/(x-\pi/2))$, and $x^2e^{\sin(1/x)}$.
Limit problems are often considered challenging when they involve one of the following forms during direct evaluation:
\[0-0,\ \frac{0}{0},\ \frac{\infty}{\infty},\ 0\cdot \infty,\ \text{and }0^0.\]To address cases of $\frac{0}{0}$ or $\frac{\infty}{\infty}$, one effective strategy is to begin by applying algebraic manipulations, including techniques such as cancellation and rationalization. (I will provide examples of these techniques in our next class.)