Search results
Results from the WOW.Com Content Network
Limits at infinity involving e. 0. Behavior of the sequences as n tends to infinity. 0. Limit infinity ...
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.
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 :
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
As far as I am aware, when doing floating point math on most computers in most computer languages, 1.0/0.0 will yield positive infinity (-1.0/0.0 yielding negative infinity). I think this question is extremely dependent on context and usage.
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 ...
$\begingroup$ 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.) $\endgroup$ –
And in the above limit in your question is exist because limit from the right equal the limit from the left equal to infinity + but if you take f(x)= 1\x in this case the limit doesn't exsist the limit from the right =infinity+ but the limit from the left equal to infinity - Please don't hesitate to discuss this with me.
$\begingroup$ Infinity is not a number. Note that even though $\lim_{x \to 0} 1/|x| = +\infty$, in common parlance, this limit does not exist, and writing that it equals $+\infty$ just gives a description of why the limit fails to exist. $\endgroup$
As Ovi noted, one theorem is that $ \lim \limits _ { x \to c } f ( g ( x ) ) = f ( w ) $ if $ w = \lim \limits _ { x \to c } g ( x ) $ exists and $ f $ is continuous there. This is completely inapplicable, since $ w $ is infinite in this case, and so there's no way that $ f $ could be continuous there; we can't even say that $ f $ is defined at ...