Need to find the limits of the following
8x-6/4x^3+6 as x approaches infinity
square root of x^4+1/x^3+4 as x approaches negative infinity
For the first one, you just simply divide both sides of the fraction by $\displaystyle x^3$:
$\displaystyle \lim_{x \to \infty} \frac{\frac{1}{x^3}}{\frac{1}{x^3}}\frac{8x - 6}{4x^3 + 6}$
This turns the limit into:
$\displaystyle \lim_{x \to \infty} \frac{\frac{8}{x^2} - \frac{6}{x^3}}{4 + \frac{6}{x^3}}$
Now we use the property that the limit of a quotient is the quotient of the limits:
$\displaystyle \frac{\lim_{x \to \infty} \frac{8}{x^2} - \frac{6}{x^3}}{\lim_{x \to \infty} 4 + \frac{6}{x^3}}$
The numerator goes to 0 and the denominator goes to 4, so that:
$\displaystyle \lim_{x \to \infty} \frac{8x - 6}{4x^3 + 6} = \frac{0}{4} = 0$
As for the second one, it involves multiplication by $\displaystyle \frac{\frac{1}{x^2}}{\frac{1}{x^2}}$, but it requires a little thought, basically that:
$\displaystyle \sqrt{\frac{1}{x^4}} = \frac{1}{x^2}$
You use that and solve it exactly as I solved the first one and you should get it.