Find the limit of $\lim\limits_{x \to+\infty} \frac{4x-3}{2x+5}$ and $\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$ Hello guys help me with these:
1.) $\lim \limits_{x \to +\infty} frac{4x-3}{2x+5}$
$=\lim \frac{4-\frac{3}{x}}{2+\frac{5}{x}}$
$= \lim \frac{4-3\frac{1}{x}}{2+5\frac{1}{x}}$ 
=$\frac{\lim 4- \lim  3  \lim \frac{1}{x}}{\lim 2+ \lim 5 \lim\frac {1}{x}}$   // lim of $\frac{1}{x}$=0
=$\frac{4 - 3 * 0}{2 + 5 * 0}$
=$\frac{4 - 0}{2 + 0}$
=$\frac{4}{2}
=2$
lim of the function as $x$ approaches positive infinity is $2$.
Is it correct?
If then what should I do if the given is:
$\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$

When you evaluate a limit approaching either positive or negative infinity, you could easily evaluate it by dividing top and bottom by $x$ which has the highest power.
$$
\lim_{x \rightarrow \infty} \frac{x}{x^2}
$$
Given / $x^2$:
$$
\lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{1} = \lim_{x \rightarrow \infty} \frac{\frac{1}{\infty}}{1} = 0 \because \frac{1}{\infty} = 0
$$
When $x$ goes to infinity only the highest power x matters. It is because how fast the value of each x changes is quite different from other's.
To take an instance:

$f(x) = x, g(x) = x^2$
When $x = 2$, $f(2) = 2$ and $g(2) = 4$
After a while, when $x = 20$, $f(20) = 20$, $g(20) = 400$

