函数的积和商的极限

There is a special property of limits pertaining to products:

$\lim_{x \rightarrow h}(f(x)g(x)) = \lim_{x \rightarrow h}(f(x)) \cdot \lim_{x \rightarrow h}(g(x))$

This also works for quotients, as a quotient is still technically a product:

$\lim_{x \rightarrow h}\frac{f(x)}{g(x)} = \lim_{x \rightarrow h}\left(f(x) \cdot \frac{1}{g(x)}\right) = \lim_{x \rightarrow h}f(x) \cdot \lim_{x \rightarrow h}\frac{1}{g(x)}$

At times this technique can make a limit easier to evaluate or at the very least make it appear less intimidating. In particular, if working with a quotient where $f(x)$ and $g(x)$ have common factors, write the limit in the form on the left-hand side of the equation so the common factors cancel.

The second option, $\lim_{x \rightarrow 0}e^x \cdot \lim_{x \rightarrow 0}x^2$, immediately follows from the initial expression by the product rule for limits.

There is a special property of limits pertaining to products:

$$\begin{equation} \lim_{x \rightarrow h}(f(x)g(x)) = \lim_{x \rightarrow h}(f(x)) \lim_{x \rightarrow h}(g(x)). \end{equation}$$

This also works for quotients, as a quotient is still technically a product:

\begin{equation} \lim_{x \rightarrow h}\frac{f(x)}{g(x)} = \lim_{x \rightarrow h}\left(f(x) \cdot \frac{1}{g(x)}\right) = \lim_{x \rightarrow h}f(x) \cdot \lim_{x \rightarrow h}\frac{1}{g(x)}, \end{equation} if $g(x) \neq 0$.

At times this technique can make a limit easier to evaluate or at the very least make it appear less intimidating. In particular, if working with a quotient where $f(x)$ and $g(x)$ have common factors, write the limit in the form on the left-hand side of the equation so the common factors cancel.

(Videos) Limit Laws to Evaluate a Limit

by PatrickJMT

Here's a video by patrickJMT where he goes over problems which can be solved using the limit laws.

Suppose that $c$ is a constant and the limits $$\lim_{x\rightarrow a}f(x)\text{ and }\lim_{x\rightarrow a}g(x)$$ exist. Then

Now that we got the boring stuff out of the way, we can begin to solve problems using these laws!

Suppose that $$\lim_{x\rightarrow2}f(x)=4\text{ and }\lim_{x\rightarrow2}g(x)=-5$$ The first problem he looks at is $$\lim_{x\rightarrow2}\frac{3f(x)-2g(x)}{f(x)}$$ In this problem we will be using the laws for division, subtraction, and constants. $$\lim_{x\rightarrow2}\frac{3f(x)-2g(x)}{f(x)}\\ =\frac{\lim\limits_{x\rightarrow2}3f(x)-2g(x)}{\lim\limits_{x\rightarrow2}f(x)}\\ =\frac{\lim\limits_{x\rightarrow2}3f(x)-\lim\limits_{x\rightarrow2}2g(x)}{\lim\limits_{x\rightarrow2}f(x)}\\ =\frac{3\lim\limits_{x\rightarrow2}f(x)-2\lim\limits_{x\rightarrow2}g(x)}{\lim\limits_{x\rightarrow2}f(x)}\\$$ Now we can plug in the limits for the functions already given to us above. $$=\frac{3\color{red}{(4)}-2\color{red}{(-5)}}{\color{red}{(4)}}\\ =\frac{12+10}{4}=\frac{22}4\\ \boxed{=\frac{11}2}$$