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## Local behaviors of a function.


**Concept here**: what does that mean by local or global.

For example, one could say 

> Around zero, the function $f(x)=x^2$ seems pertty similar to $f(x)=x^3+x^2$

When I say local behavior, I just mean a very, very tiny neigborhood around that point. 

## Now think about why the local behavior of $f(x)=x^2$ and $f(x)=x^2+x^3$ are similar around $x=0$

Hint: $x^2+x^3=(1+x)x^2$. Thinking by this way: $1+x$ is almost equal to 1 around $x=0$.

### What is happening

around $x=0$, the term $x^2$ is very very small. BUT, at the same time, $1+x$ is not small, $1+x$ is just around $1$

Think about what is happening if we multiply a tiny number with a number that is close to 1.

$$
0.00001\times 1.000001=0.00001000001
$$

<span style="color:red">**Key point: if we multiply a number close to 1 with a number close to 0. The result would be almost determined by the number close to 0**</span>

**0 contributes more than non-zero to local behaviors**

### Now think about local behavior of $f(x)=\frac {1+x}{x^2}$ and $f(x)=\frac1{x^2}$ around $x=0$

Why the local behavior of this two function around $0$ are the same? it is beause, around $x=0$, $1+x$ is amost equal to 1, and the main contribution of the local behavior is contributed by $\frac1{x^2}$

$$
\lim_{x\rightarrow 0}\frac1{x^2}=\infty
$$


## Now let's think about how to use local behaviors to help us draw graph of functions.

#### Example: Draw the graph of 
$$
f(x)=\frac{x}{(x+1)(x-1)}
$$

**Key point: analyze its local behavior at some special point**

What is special point?
> It is some point where the function equals to 0 or $\infty$

Which 3 points is the point that the function is 0 or $\infty$?

$x=0,x=1,x=-1$

Try to analyze the local behavior of it.

When $x=-1$, we factorize the function into the following part

$$
f(x)=\underbrace{\frac1{x+1}}_{\text{goes to }\infty}\cdot \underbrace{\frac x{x-1}}_{\text{around}\frac12}
$$

Then the fuction
$$
g_{-1}(x)=\frac12 \cdot\frac1{x+1}
$$
has the same local behavior as $f(x)$ around $x=-1$

**Excercise**: Analyze local behaviros of $f$ around $x=0$ and $x=1$

When $x=0$,
$$
f(x)=\underbrace{x}_{\text{goes to 0}}\cdot\underbrace{\frac1{(x+1)(x-1)}}_{\text{around}-1}
$$
$$
g_0(x)=-x
$$

Left the case of $x=1$ for your own excercise

$$
g_1(x)=\frac12\cdot\frac{1}{x-1}
$$

### Example

$$
f(x)=\frac{(x+1)^2(x+2)^3}x
$$

### The most important point is 
$$
x=0,x=-1,x=-2
$$
Where when we considering around $x=0$, we want to factorize this function into the following parts
$$
f(x)=\frac1x\cdot(x+1)^2(x+2)^3
$$

$$
g_0(x)=\frac8x
$$

When we considering around $x=-1$, we want to factorize this function into the following parts

$$
f(x)=(x+1)^2\cdot \frac{(x+2)^3}x
$$

$$
g_{-1}(x)=(x+1)^2\cdot(-1)
$$

When we considering around $x=-2$,


$$
f(x)=(x+2)^3\cdot\frac{(x+1)^2}x
$$

$$
g_{-2}(x)=(x+2)^3\cdot \left(-\frac12\right)
$$


How to determine the bahavior around $\infty$

For a polynomial, the behavior of the function when $x$ goes to infty, is given by the highest term

$$
f(x)=x^3+3x^2+x+7
$$

$$
g_{\infty}(x)=x^3
$$

How about rational functions?

The behavior of the function when $x\rightarrow \infty$ is given by the ratio of the highest term on top and bottom

$$
f(x)=\frac{3x^3+2x^2+1}{2x^2+1}
$$

$$
g_\infty(x)=\frac{3x^3}{2x^2}=\frac32 x
$$
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