Help with limit equation The question is:
$$\lim_{x \to + \infty} \frac{3x^4 - 11x^3 + 42x}{-15x^4 + 3x^2 +101}$$
So far my answer:
$$\lim_{x \to + \infty} \frac{x^4}{x^4} \frac{3 - \frac{11x^3}{x^4}+\frac{42x}{x^4}}{-15 + \frac{3x^2}{x^4}+ \frac{101}{x^4}}$$
cancel the $x^4$ gives us:
$$\lim_{x \to + \infty} \frac{3 - \frac{11x^3}{x^4}+\frac{42x}{x^4}}{-15 + \frac{3x^2}{x^4}+ \frac{101}{x^4}}$$
Now as x becomes larger (as  $x \to +\infty$) $\frac{1}{x^4}$ goes to $0$
this gives:
$$\lim_{x \to + \infty}  \frac{3 - 11\cdot 0 + 42 \cdot 0}{-15 + 3\cdot 0 + 101 \cdot 0}$$
$$=- \frac{3}{15}$$
I have been following my text book so this is how I got to where I'm at I was just wondering if I have gone about this the right way?

Yes. It is right. When you have ratios of polynomials, however, there is an easy way to analyze $$\lim_{x \to +\infty} \frac{p(x)}{q(x)}$$
It is $+\infty$ or $-\infty$ if the degree of $p(x)$ is greater than the degree of $q(x)$, and the sign depends on the sign of the ratio of $p(x)$ and $q(x)$'s dominant coefficients. It is zero is the degree of $p(x)$ is less than the degree of $q(x)$. If both degrees are equal, the limit is the ratio of the dominant coefficients. Your exercise is an example of the last case.
