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

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