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,

1. $\displaystyle\lim_{x\to a} f(x)\pm g(x) = \lim_{x\to a}$
2. $\displaystyle\lim_{x\to a} f(x)g(x) = \lim_{x\to a}f(x)\lim_{x\to a}g(x)$
3. $\displaystyle\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ if $\displaystyle\lim_{x\to a}g(x)\neq 0$
4. $\displaystyle\lim_{x\to a} (f(x))^{(b/c)} = \left(\lim_{x\to a}f(x)\right)^{(b/c)}$ if $c$ is odd and $b/c$ is in its lowest terms (最簡分數形式).

Algebraic tricks

Kill a common zero and the rationalization

There are two scenarios where you might want to consider employing this technique. The first scenario arises when you encounter an indeterminate form, specifically $\frac{0}{0}$, and the denominator takes the form of $\sqrt{\cdot}-\sqrt{\cdot}$ or $1-\cos(x)$. The second scenario occurs when you are working with a limit as $x$ approaches infinity, and direct evaluation leads to the indeterminate form $\infty - \infty$. These are two common situations where the technique we’ll discuss becomes particularly useful.

To illustrate these scenarios, let’s explore several typical examples.

Example 2. $\displaystyle\lim_{x\to\infty}\sqrt{x}-\sqrt{x+1}$

Example 3. $\displaystyle\lim_{x\to 0}\frac{x^2}{1-\cos(x)}$

Help from some well-known limits

The following limits are useful when we tried to evaluate other limits:

• $\displaystyle\lim_{x\to 0}\frac{\sin(x)}{x}=1$.

Example. $\displaystyle\lim_{x\to 0}\frac{\sin(2x)}{\sin(3x)} = \lim_{x\to 0}\frac{\sin(2x)\cdot 3x}{2x\cdot\sin(3x)}\cdot\frac{3x}{2x} = \frac{3}{2}$.

Change of variables

A change of variables allows us to simplify the expression of a limit, making it easier to evaluate. One of the most commonly used changes of variables is $t = \frac{1}{x}$. Using this substitution, we can rewrite a limit, such as $\lim_{x\to \infty}\sin\left(\frac{1}{x}\right)^{\ln\left(\frac{1}{x}\right)}$, as $\lim_{t\to 0}\sin(t)^{\ln(t)}$. This new form is much simpler to apply L’Hôpital’s rule to.

Continuous functions

If $f(x)$ is a continuous function, then we have

This theorem enables us to temporarily disregard the outermost continuous function and concentrate on evaluating the limit based on the inner function.

The Squeeze theorem

Theorem. Let $I$ be an interval containing the point $a$. Let $g$, $f$, and $h$ be functions defined on $I$, except possibly at $a$ itself. Suppose that for every $x$ in $I$ not equal to $a$, we have \(g(x)\leq f(x)\leq h(x)\) and also suppose that \(\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L.\) Then $\displaystyle\lim_{x\to a}f(x)=L$.

Indeterminate form

We refer to a limit as having an ‘indeterminate form’ when it cannot be straightforwardly determined by direct evaluation. The following are most common indeterminate forms:

L’Hôpital’s Rule

Theorem(L’Hôpital’s rule) Let $f(x)$ and $g(x)$ be differentiable at $x=a$, and $g’(x)\neq 0$ for some interval containing $a$. If the limit $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}$ is indeterminate form and $\displaystyle\lim_{x\to a}\frac{f’(x)}{g’(x)}$ is not DNE or is equal to $\pm\infty$, then \(\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}.\)

Remark. $x\to a$ can be changed to $x\to a^+$, $x\to a^-$, $x\to \infty$, or $x\to -\infty$.

We give two examples when L’Hopital rule is not applicable.

Example: The limit $\lim_{x\to 0}\frac{x+1}{2x+1}$ is not an indeterminate form, so L’Hôpital’s Rule does not apply. Therefore, we cannot use L’Hôpital’s Rule here. The limit evaluates to $\frac{1}{2}$.

Example: Although the limit $\lim_{x\to \infty}\frac{x-\cos(x)}{x-\sin(x)}$ is initially of an indeterminate form, we cannot use L’Hôpital’s Rule directly. The actual limit is $\lim_{x\to\infty}\frac{1-\frac{\cos(x)}{x}}{1+\frac{\sin(x)}{x}}=1$. L’Hôpital’s Rule is not applicable here, as it would lead to an incorrect result of “DNE” (Does Not Exist).