# The limitation of a function

> Graph math problem will give you a better experience of solving problems. Imaging too. But so sad that I can't draw graphs for this book on this period of time.

$$
\lim\_{x \to \infty}{\frac{1}{x^2}} = 0
$$

Why does this happen? We could imagine, if $$x^2$$ goes on and on, to infinite, $$\frac{1}{x^2}$$ will more and more approaching $$0$$.

There's another limit equation that similar to the above:

$$
\lim\_{x \to 0}{\frac{1}{x^7}} = \infty
$$

Why does this happen? If $$x^7$$ approach to $$0$$, $$\frac{1}{x^7}$$ goes bigger and bigger, it's infinite.

As you can see, normally, if you are asked getting the limitation of a basic simple function, all you have to do is calculate the y value of f(x).

$$
\begin{align\*}
\lim\_{x \to 1}{x^2 + 5} &= 6
\ \\
\lim\_{x \to -2}{\sin(x + 2)} &= 0
\ \\
\lim\_{x \to 1}{\sqrt\[999]{x^{277}}} &= 1
\end{align\*}
$$

![](https://2889346953-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-Le4Kw7maZEabKFjsC7z%2F-LfP6qH5Be5If0f2YPUW%2F-LfP7-cp1hzKTJiwt8PN%2Fx%5E2.png?generation=1558437695279431\&alt=media)

Above is a graph of $$f(x) = x^2$$.

If I ask you what's the limitation of $$\lim\_{x \to 2}{x^2}$$, the real meaning is $$\lim\_{x \to 2^-}{x^2}$$ and $$\lim\_{x \to 2^+}{x^2}$$ **exist and equal.**

$$\lim\_{x \to 2^-}{x^2}$$ means `x` from `less than 2` to `2`, what `y` will be approaching

$$\lim\_{x \to 2^+}{x^2}$$ means `x` form `greater than 2` to `2`, what `y` will be approaching to.

Here comes a new definition of continuity when you think about it. No matter greater than 2 or less than 2, they all approaching 4, this means they are continuous in x


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