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.

limx1x2=0\lim_{x \to \infty}{\frac{1}{x^2}} = 0

Why does this happen? We could imagine, if x2x^2 goes on and on, to infinite, 1x2\frac{1}{x^2} will more and more approaching 00.

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

limx01x7=\lim_{x \to 0}{\frac{1}{x^7}} = \infty

Why does this happen? If x7x^7 approach to 00, 1x7\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).

limx1x2+5=6limx2sin(x+2)=0limx1x277999=1\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)=x2f(x) = x^2.

If I ask you what's the limitation of limx2x2\lim_{x \to 2}{x^2}, the real meaning is limx2x2\lim_{x \to 2^-}{x^2} and limx2+x2\lim_{x \to 2^+}{x^2} exist and equal.

limx2x2\lim_{x \to 2^-}{x^2} means x from less than 2 to 2, what y will be approaching

limx2+x2\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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