# Generalization of limx→±âˆ‍\lim_{x\to ±âˆ‍}

I have a question about the generalization of \lim_{x\to آ±âˆ‍}\lim_{x\to آ±âˆ‍}.My teacher gave us 3 rules to deal with this. 1)1)If the highest powers are the same on the numerator and denominator, then just take the ratio of the highest powers. For example in \lim_{x\to âˆ‍}(\frac{2x^2-3}{x^2+9}\lim_{x\to âˆ‍}(\frac{2x^2-3}{x^2+9}), the answer is 2. 2)2) If the denominator has a higher power, then the answer is always 00. For example in \lim_{x\to آ±âˆ‍}(\frac{x}{x^5+1})\lim_{x\to آ±âˆ‍}(\frac{x}{x^5+1}), the answer is automatically 00. 3)3) If the power of the numerator is higher, then then take the ratio of the numerator to the denominator of the highest power. For example in \lim_{x\to -âˆ‍}(\frac{7x^2+x+11}{4-x})\lim_{x\to -âˆ‍}(\frac{7x^2+x+11}{4-x}), the ratio is \frac{7x^2}{-x}=-7x\frac{7x^2}{-x}=-7x.Plugging in -âˆ‍, it follows that the limit approaches infinity. As shown by these examples, the method definitely works. It also has some logic behind it. The logic behind rule (1)(1) is that the infinities “cancel” each other out, just leaving a ratio.The logic behind rule (2)(2) is that if the numerator is a lower power, the infinity on the bottom will always cause the answer to be 00. The logic behind rule (3)(3) is that yet again, the infinities “cancel” each other out, leaving a ratio. So, my question is why don’t other schools teach it like this? And why isn’t it in my calc textbook? It works almost the same as the method the textbook uses, but it’s way faster.

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Other than the “logic” you mention, those are standard rules we teach in freshmen calculus.
– Bye_World
Oct 21 at 1:39

@Bye_World My textbook does it differently though, it usually says to divide by the highest power to let the terms cancel out when you take the limit. It takes a lot longer…
– Aaron M
Oct 21 at 1:43

That’s how you derive your rules. (And it doesn’t take that much longer ;p)
– Bye_World
Oct 21 at 1:44

The textbook is probably just being more rigorous about it. Presumably you have a theorem which says that \lim_{x \to \infty}1/x = 0\lim_{x \to \infty}1/x = 0. By dividing the numerator and denominator by the highest power you can usually get the expression in terms of a bunch of 1/x1/x expressions and then invoke the theorem along with standard limit laws to get the answer.
– wgrenard
Oct 21 at 1:46

Yeah @wgrenard that’s exactly the method they use.
– Aaron M
Oct 21 at 2:11

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