I am currently taking a course on Calculus 1. We just recently went over L'Hopital's Rule, which states the following:
When evaluating $\lim\limits_{x \to a} \frac{f(x)}{g(x)}$, if direct substitution yields 0/0 or $\pm\frac{\infty}{\infty}$, then the limit of our original expression is equal to $\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$.
However, I recently came across the following problem:
$\lim\limits_{x \to \infty} \frac{5x – 2x^2}{e^x+3}$
While the solution key indicates that we may simply use L'Hopital's Rule, I do not understand why. Moreover, if we utilize direct substitution, we will be left with $\frac{\infty – \infty}{\infty}$. Presently, this is not equivalent to the indeterminate form $\pm\frac{\infty}{\infty}$. Is there some sort of manipulation we can do to show that $\frac{\infty – \infty}{\infty}$ is equivalent to $\pm\frac{\infty}{\infty}$?
Best Answer
As $x$ becomes very large, $|-2x^2|$ is much larger than $|5x|$. So the top is "equal" to $-\infty$, because the $5x$ term is negligible in size compared to the $-2x^2$ term. (For example, if we plug in $x = 1000$, we get $-2\cdot(1000)^2 + 5\cdot100 = -2000000 + 5000$.)
In general, any non-constant polynomial will be plus or minus infinity for the purposes of L'Hopital's Rule. (Positive infinity if the largest power term has a positive coefficient, minus if the largest power term has a negative coefficient.) And as other commenters mention, you don't actually need the top to be infinity to apply L'Hopital's rule; it works either way.
It's worth being very careful when you substitute infinity into these expressions directly. Infinity isn't a real number that you can add and subtract; the direct substitution method is a shorthand for finding the limit of the top and the bottom. If you're confused, try to find the limit directly.