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.


This is because when you wrtie


not only is this mathematically incorrect, but you're losing valuable information about the limit. In this case,


So by ignoring the sign on the expression, and just writing

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


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