# Question about the rules to follow when evaluating infinite limits

I learned to solve infinite limits diving the whole function by the leading term, knowing that when you have a fraction, if it is in the numerator you will get $\infty$, and if it is in the denominator a number.

For example, this is how I would solve this exercise:

$$\lim _{x\to \infty }\left(\frac{x^4+2}{3x+5}\right)=\lim _{x\to \infty }\left(\frac{1+\frac{2}{x^4}}{\frac{3}{x^3}+\frac{5}{x^4}}\right)=\lim _{x\to \infty }\left(\frac{1+0}{0+0}\right)=\frac{1}{0}=\infty$$

However, there seems to be something that my teacher forgot to tell me, and I could not find anything about that on the internet. I noticed there is something wrong when trying to solve the following exercise:

$$\lim _{x\to \infty }\left(\frac{7x^5+1}{-3x+8}\right)=\lim _{x\to \infty }\left(\frac{7+\frac{1}{x^5}}{-\frac{3}{x^4}+\frac{8}{x^5}}\right)=\lim _{x\to \infty }\left(\frac{7+0}{-0+0}\right)=\frac{7}{0}=\infty$$

The answer is $\infty$, but the sign should be negative: $-\infty$. When I tried to evaluate the limit with an online tool, it says that you divide the function by the highest denominator power, doing this instead:

$$\lim _{x\to \infty }\left(\frac{7x^5+1}{-3x+8}\right)=\lim _{x\to \infty }\left(\frac{7x^4+\frac{1}{x}}{-3+\frac{8}{x}}\right)=\lim _{x\to \infty }\left(\frac{7x^4}{-3}\right)=\frac{\infty \:}{-3}=-\infty$$

Why should I choose the highest denominator power in this case? What makes it different from the first example I wrote? Because if I try to evaluate other functions following this rule I get the wrong answers, so it does not always work.

$$\frac{7}{0}$$
$$-0=0$$
$$\frac{7}{0}$$ you were actually missing the fact that the limit goes to $-\infty$ and not $\infty$