2017年12月25日 · When we use straightforward approach, we get $$ \frac{\infty+1}{\infty} = \frac{\infty}{\infty} $$ In the process of investigating a limit, we know that both the numerator and denominator are going to infinity.. but we dont know the behaviour of each dynamics. But if we investigate further we get : $$ 1 + \frac{1}{x} $$ Some other examples :

2016年4月28日 · First of all: you cannot just subtract infinity from infinity. Infinity is not a real number so you can't simply use the basic operations as you're used to do with (real) real numbers. However, in the context of limits, there are things you can do.

2020年2月14日 · Because of this theorem, one might argue that it is fair to "split the limits", as you say, resulting in the "infinity arithmetic" expression. $$\infty + \infty = \infty $$ Fine so far. But just because one can write an "infinity arithmetic" expression does not mean there is a theorem supporting that expression.

2012年8月11日 · Essentially, you gave the answer yourself: "infinity over infinity" is not defined just because it should be the result of limiting processes of different nature. I.e., since such a definition would be given for the sake of completeness and coherence with the fact "the limiting ratio is the ratio of the limits", your

2015年9月11日 · On the contrary, those limits tell you that the limit of the entire quotient is $0$. This may be easier to see if you rewrite to $$ \lim_{x\to\infty} f(x)\frac1{h(x)} $$ where $\lim_{x\to\infty} f(x) = 0 $ and $\lim_{x\to\infty} \frac1{h(x)}=0 $, and the product of two functions that both have limit $0$ surely also has limit $0$.

2016年10月27日 · $\begingroup$ This works when the limits both exist, since $\exp$ and $\log$ are both continuous. (Phrase $\lim r^s$ as $\lim \exp(s \log r)$, and use that the limit of a product is the product of the limits.) $\endgroup$ –

Evaluating Limits at Infinity. 2. Limit as x approches infinity - Trouble with calculus. 0.

Your title says something else than "infinity times zero". It says "infinity to the zeroth power". It is also an indefinite form because $$\infty^0 = \exp(0\log \infty) $$ but $\log\infty=\infty$, so the argument of the exponential is the indeterminate form …

2018年6月6日 · I know that indeterminate forms exist in limits, such as $\\frac{0}{0}$, $\\frac{\\infty}{\\infty}$, $0^0$, $\\infty^0$, $1^\\infty$. Then, if $\\lim\\limits_{x \\to ...

I'm working with limits at infinity and stumbled upon this exercise where I want to evaluate the indicated limit: $$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x}$$ I tried to solve it by doing the