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:
$\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$

$$\lim_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1} =\lim_{x \to -\infty}  \frac{2x^2(1-\frac{1}{2x}+\frac{5}{2x^2})}{4x^3(1-\frac{1}{4x^3})} =\lim_{x \to -\infty}  \frac{(1-\frac{1}{2x}+\frac{5}{2x^2})}{2x(1-\frac{1}{4x^3})}$$
It is clear that 
$$\lim_{x \to -\infty} \frac{1}{4x^3} =\lim_{x \to -\infty} \frac{1}{2x} = \lim_{x \to -\infty} \frac{5}{2x^2}= 0,$$
and so
$$\lim_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1} = \lim_{x \to -\infty} \frac{1}{2x} = 0.$$
Note: Using the same process it easy to observe that
$$\lim_{x \to -\infty} \frac{2x^3-x+5}{4x^3-1} = \lim_{x \to -\infty}  \frac{2x^3(1-\frac{1}{2x^2}+\frac{5}{2x^3})}{4x^3(1-\frac{1}{4x^3})} = \lim_{x \to -\infty}  \frac{(1-\frac{1}{2x^2}+\frac{5}{2x^3})}{2(1-\frac{1}{4x^3})} = \frac{1}{2}$$
