Limit of rational function solving I have to solve limit of rational function, but it turns out I do mistake somewhere. Where I do wrong? Does my calculations correct?
$$\lim_{x\to \infty}\frac{x^3-2x-1}{x^5-2x-1}$$
Step 1: 
$$\lim_{x\to\infty}\frac{x^5\left(\frac{1}{x^2}-\frac{2}{x^4}-\frac{1}{x^5}\right)}{x^5\left(1-\frac{2}{x^4}-\frac{1}{x^5}\right)}$$
Step 2: 
$$\lim_{x\to\infty}\frac{x^5\left(\frac{1}{x^2}-\frac{2}{x^4}-\frac{1}{x^5}\right)}{x^5\left(1-\frac{2}{x^4}-\frac{1}{x^5}\right)} = \frac{0-0-0}{1-0-0}=\frac{0}{1}=0$$
 A: The post edit answer is correct. However, it can be obtained with little (or practically no) calculations at all by using the properties of end behaviour of polynomials. 
The end behaviour of a polynomial function can be defined, informally, as the behaviour of the function's graph as it approaches $\pm \infty$. 
For polynomial functions, the degree and the leading coefficient determine the end behaviour of the function. That means, as $x$ tends to infinity it is enough to analyse the leading term and coefficient to have an idea of what the graph will look like. 
Limits are, on one hand, asking about the end behaviour. In your case, you have two polynomial functions that make up a rational function. Therefore we can combine the property of end behaviour of polynomials and the quotient rule of limits, $${\lim\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} }={ \frac{{\lim\limits_{x \to a} f\left( x \right)}}{{\lim\limits_{x \to a} g\left( x \right)}},\;\;\;}\kern-0.3pt 
{\text{if}\;\;\lim\limits_{x \to a} g\left( x \right) \ne 0}$$ to quickly get the answer by considering $$\lim_{x\to \infty} \frac{x^3}{x^5} = \lim_{x \to \infty} \frac{1}{x^2} = \frac{1}{\infty} = 0$$ Of course, all this can be done in the head in just a matter of seconds.
To Sum It Up
Suppose you are given two rational functions $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$and $$g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$$
Then the value of $$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$$ can be determined as follows:
$\bullet$ If $n>m$  and the sign of $a_n$ is the same as that of $b_m$ (i.e. $\frac{a_{n}}{b_{m}}$ is positive) then the limit is $\infty$
$\bullet$ If $n>m$  and the sign of $a_n$ is different from that of $b_m$, (i.e. $\frac{a_{n}}{b_{m}}$ is negative) then the limit is $-\infty$
$\bullet$ If $n<m$ then the limit is $0$.
$\bullet$ If $n=m$ then the limit is the value of $\frac{a_{n}}{b_{m}}$.
Note that this method is only for limits to infinity
