What is the limit of $f(x)=\lim_\limits{n\to \infty}\frac{x^n}{x^n+3}$ when $x$ tends to a certain point? Let: $f(x)=\lim_\limits{n\to \infty}\frac{x^n}{x^n+3}.$
I need to calculate:


*

*$\lim_{x \rightarrow 1^{+}} f(x)$

*$\lim_{x \rightarrow 1^{-}} f(x)$

*$\lim_{x \rightarrow (-1)^{+}} f(x)$

*$\lim_{x \rightarrow (-1)^{-}} f(x)$


I have no idea how to deal with this kind of questions. I'd like to some directing or hints. 
 A: Notice:


*

*$$\lim_{n\to\infty}\space\frac{x^n}{x^n+3}=\lim_{n\to\infty}\space\frac{1}{1+\frac{3}{x^n}}=0\space\space\space\space\space\space\text{if}\space\space|x|<1$$

*$$\lim_{n\to\infty}\space\frac{x^n}{x^n+3}=\lim_{n\to\infty}\space\frac{1}{1+\frac{3}{x^n}}=1\space\space\space\space\space\space\text{if}\space\space x<-1$$

*$$\lim_{n\to\infty}\space\frac{x^n}{x^n+3}=\lim_{n\to\infty}\space\frac{1}{1+\frac{3}{x^n}}=1\space\space\space\space\space\space\text{if}\space\space x>1$$
A: Hint:
Use the fact that:


*

*For $|x|<1$ : $x^n =0$ for $n \rightarrow \infty$

*For $1<|x|$: $x^n \rightarrow \infty$ for $n \rightarrow \infty$
So for $|x|>1$, the 3 in the denominator becomes negligible as compared to $x^n$ and hence you the ratio as 1, whereas for $|x|<1$ you have $x^n$ tending to 0 which makes the $x^n$ in the denominator negligible as compared to 3 whereas the numerator tends to 0. Hence, you have 0 as the limit (since you have $\lim_\limits{h \rightarrow 0}\frac {h}{h+3}$).
A: First, let's know what $f(x)$ equals to.


*

*If $|x| < 1$, then 


$$\lim_{n \to \infty} \frac{x^n}{x^n + 3} = 0$$


*

*If $|x| > 1$, then


$$\lim_{n \to \infty} \frac{x^n}{x^n + 3} = 1$$
Therefore:
$$f(x) = \begin{cases} 0, x \in (-1,1) \\ 1, x \in (-\infty, -1)\cup (1,\infty) \end{cases}$$
Now you can calculate the limits directly.
