总结：找极限的方法

Calculating Limits Using Limit Laws

Assume the limits $\lim_{x→c} f(x)$ and $\lim_{x→c} g(x)$ exist. Let $a$ be any number.

1. Sum law: $\lim_{x→c}(f(x) + g(x)) = \lim_{x→c}f(x) + \lim_{x→c}g(x)$

2. Difference law: $\lim_{x→c}(f(x) - g(x)) = \lim_{x→c}f(x) - \lim_{x→c}g(x)$

3. Constant multiple law: $\lim_{x→c}(af(x)) = a\lim_{x→c}f(x)$

4. Product law: $\lim_{x→c}(f(x)g(x)) = (\lim_{x→c} f(x)) (\lim_{x→c} g(x))$

5. Quotient law: if $\lim_{x→c}g(x) \ne 0$ then $lim_{x→c} \frac{f(x)}{g(x)} = \frac{\lim_{x→c}f(x)}{\lim_{x→c}g(x)}$

6. Power law: $\lim_{x→c}(f(x))^n = (\lim_{x→c}f(x))^n$

7. $\lim_{x→c}a = a$

8. $\lim_{x→c}x = c$

9. $\lim_{x→c}x^n = c^n$ where $n>0$

10. Root Law: $\lim_{x→c}\sqrt[n]{f(x)} =\sqrt[n]{\lim_{x→c}f(x)}$ where $n > 0$ and if $n$ is even we assume $f(x) \ge 0$

We can use these laws to numerically estimate the limits of functions.

Example 1:

$$\lim_{x->2}\frac{3x-4}{x^2+5}$$

Using the Difference Law we get: $$= \frac{\lim_{x->2}3x-4}{\lim_{x->2}x^2+5}$$ Using the Sum, Difference and Constant Multiple Law we get:

$$= \frac{3\lim_{x->2}x - lim_{x->2}4}{\lim_{x->2}x^2+\lim_{x->2}5}$$

We evaluate this limit using the 7th, 8th and 9th Law listed above:

$$= \frac{3 \cdot 2 - 4}{2^2 + 5}$$ $$=\frac{2}{9}$$

Note: You can solve limits by plugging $c$ in place of $x$ however, there are cases when you will get an undefined value. If that happens you must manipulate the function before plugging in c or using the limit laws.

Example 2:

$$\lim_{x->2}\frac{\sqrt{x^2+4}}{\sqrt{x^2-4}+2}$$

Here we see that if we plug in 2 the denominator equals 0 which gives us an undefined value. We can manipulate this function to get the correct limit. First, multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of a + b is a - b. The conjugate of a - b is a + b.

$$= \lim_{x->2}\frac{\sqrt{x^2+4}}{\sqrt{x^2-4}+2} \cdot \frac{\sqrt{x^2-4}-2}{\sqrt{x^2-4}-2}$$ $$= \lim_{x->2}\frac{\sqrt{x^2+4}(\sqrt{x^2-4}-2)}{x^2-4 -4}$$

Now we can use limit the limit laws or plug in $c$

$$=\frac{\sqrt{8}(0 - 2)}{4 - 4 - 4}$$ $$=\frac{2\sqrt{2}(-2)}{- 4}$$ $$=\sqrt{2}$$

Test Yourself!

(Video) Calculating a Limit by Factoring and Canceling

by PatrickJMT

Here's a video by patrickJMT showing you how to find a limit of a function using the factor canceling technique.

He demonstrates this technique by looking at an example: $$\lim_{x\rightarrow2}\frac{x^2-4}{x-2}$$ Notice what happens when we plug in 2 straight away. $$\begin{align} \frac{x^2-4}{x-2}&=\frac{\color{red}{(2)}^2-4}{\color{red}{(2)}-2}\\ &=\frac{4-4}{2-2}\\&=\frac00 \end{align}$$ So we have to be creative and figure something else out. We can instead factor this expression. $$\lim_{x\rightarrow2}\frac{x^2-4}{x-2}=\lim_{x\rightarrow2}\frac{(x+2)(x-2)}{x-2}$$ When we factor the top we get an $x-2$ in both the numerator and denominator. This can be canceled out. $$\lim_{x\rightarrow2}\frac{(x+2)\color{red}{(x-2)}}{\color{red}{x-2}}=\lim_{x\rightarrow2}x+2$$ Now that it's simplified let's try plugging 2 back in. $$x+2=\color{red}{(2)}+2=4$$ So $$\lim_{x\rightarrow2}\frac{x^2-4}{x-2}=4$$

If we graph the function $y=\dfrac{x^2-4}{x-2}$ it's going to look a lot like the function $y=x+2$ except it will be undefined at $x=2$ because we have a zero in the denominator.

As you can see when we put in points getting closer and closer to 2, the function gets closer and closer to 4.

(Video) Calculating a Limit by Expanding and Simplifying

by PatrickJMT

Here's a video by patrickJMT showing you how to find a limit of a function using the expanding and simplifying technique.

He demonstrates this with the example: $$\lim_{h\rightarrow0}\frac{(4+h)^2-16}{h}$$ Notice what happens when we plug in $h=0$. $$\begin{align} \frac{(4+h)^2-16}{h}&=\frac{(4+\color{red}{(0)})^2-16}{\color{red}{(0)}}\\ &=\frac{4^2-16}{0}\\ &=\frac00 \end{align}$$ So we have to do something algebraic in order to figure out this limit. Notice that there's this $(4+h)^2$ that can be expanded, so let's do that. $$\begin{align} &\lim_{h\rightarrow0}\frac{(4+h)^2-16}{h}\\ &=\lim_{h\rightarrow0}\frac{16+8h+h^2-16}{h} \end{align}$$ The 16's will cancel out and then we can factor out an $h$ from the numerator and denominator. $$\begin{align} &\lim_{h\rightarrow0}\frac{16+8h+h^2-16}{h}\\ &=\lim_{h\rightarrow0}\frac{8h+h^2}{h}\\ &=\lim_{h\rightarrow0}8+h \end{align}$$ Now we can plug in $h=0$. $$8+h=8+\color{red}{(0)}=8$$ So $$\lim_{h\rightarrow0}\frac{(4+h)^2-16}{h}=8$$

(Video) Calculating a Limit by Getting a Common Denominator

by PatrickJMT

Here's a video by patrickJMT showing you how to find the limit of a function using the common denominator technique. Note: he solves this problem in a slightly different way than I do. Both get the same result.

The example he looks at is: $$\lim_{h\rightarrow0}\frac{(3+h)^{-1}-3^{-1}}{h}$$ Notice what happens when we plug in $h=0$. $$\begin{align} \frac{(3+h)^{-1}-3^{-1}}{h}&=\frac{(3+\color{red}{(0)})^{-1}-3^{-1}}{\color{red}{(0)}}\\ &=\frac{3^{-1}-3^{-1}}{0}\\ &=\frac00 \end{align}$$ Since this doesn't work we have to figure out a way to rearrange it until it does work. One thing to note is we have terms raised to the negative power. We can rewrite them in the denominator with a positive power. Be careful though, they're combined with a negative sign so we can't just put them in the denominator. We have to break the fraction up first. $$\begin{align} &\lim_{h\rightarrow0}\frac{(3+h)^{-1}-3^{-1}}{h}\\ &=\lim_{h\rightarrow0}\frac{(3+h)^{-1}}{h}-\frac{3^{-1}}{h} \end{align}$$ Now we can move the terms to the denominator. $$\begin{align} &\lim_{h\rightarrow0}\frac{(3+h)^{-1}}{h}-\frac{3^{-1}}{h}\\ &=\lim_{h\rightarrow0}\frac{1}{h(3+h)}-\frac{1}{3h} \end{align}$$ We can now find a common denominator for these fractions in order to combine them. $$\begin{align} &\lim_{h\rightarrow0}\frac{1}{h(3+h)}-\frac{1}{3h}\\ &=\lim_{h\rightarrow0}\frac{1}{h(3+h)}\color{red}{\cdot\frac33}-\frac{1}{3h}\color{red}{\cdot\frac{3+h}{3+h}}\\ &=\lim_{h\rightarrow0}\frac{3}{3h(3+h)}-\frac{3+h}{3h(3+h)}\\ &=\lim_{h\rightarrow0}\frac{3-(3+h)}{3h(3+h)}\\ &=\lim_{h\rightarrow0}\frac{-h}{3h(3+h)}\\ &=\lim_{h\rightarrow0}-\frac{1}{3(3+h)}\\ &=\lim_{h\rightarrow0}-\frac{1}{3h+9} \end{align}$$ Now we can plug in $h=0$. $$\begin{align} -\frac{1}{3h+9}&=-\frac{1}{3\color{red}{(0)}+9}\\ &=-\frac19 \end{align}$$ So $$\lim_{h\rightarrow0}\frac{(3+h)^{-1}-3^{-1}}{h}=-\frac19$$ Usually, when there's fractions lying about, it's a good idea to find common denominators and combine.

(Videos) Calculating a Limit by Multiplying by a Conjugate

by PatrickJMT

Here's a video by patrickJMT showing you how to find the limit of a function using the conjugate technique.

He shows how to use this technique by looking at an example. $$\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}$$ Notice what happens when we plug in 7. $$\begin{align} \frac{\sqrt{x+2}-3}{x-7}&=\frac{\sqrt{\color{red}{(7)}+2}-3}{\color{red}{(7)}-7}\\ &=\frac{3-3}{7-7}\\ &=\frac00 \end{align}$$ We have to do something to this in order to change it. The key things to spot are that there's a radical and two terms in the numerator. A common trick when we have a radical is to multiply by the conjugate.

Remember that if you have $a\pm b$ the conjugate is the same thing but the sign is reversed: $a\mp b$.

So let's multiply the top and bottom by the conjugate of the numerator. $$\begin{align} &\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}\\ &=\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}\color{red}{\cdot\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}}\\ &=\lim_{x\rightarrow7}\frac{(x+2)-9}{(x-7)(\sqrt{x+2}+3)}\\ &=\lim_{x\rightarrow7}\frac{x-7}{(x-7)(\sqrt{x+2}+3)}\\ &=\lim_{x\rightarrow7}\frac{1}{\sqrt{x+2}+3} \end{align}$$ Now we can plug in $x=7$. $$\begin{align} \frac{1}{\sqrt{x+2}+3}&=\frac{1}{\sqrt{\color{red}{(7)}+2}+3}\\ &=\frac1{3+3}\\ &=\frac16 \end{align}$$

So $$\lim_{x\rightarrow7}\frac{\sqrt{x+2}-3}{x-7}=\frac16$$

Summary of Techniques for Evaluating Limits

In the simplest cases we can evaluate limits by substitution. Other limits, like when we compute derivatives at a point later on, might require a little algebraic manipulation first (usually factoring and "canceling" common factors). Limit rules (such as limits of sums, differences, and constant multiples) and the squeeze theorem can sometimes help too.