Is the limit of any integer divided by infinity 0? (calculus)?

Question

I understand that 1 divided by infinity equals 0, but is this also true for any other integer? For example, does 17 divided by infinity equal 0? For that matter, does -17 divided by infinity equal 0?

Thanks in advance!

Answer 1

Your question is a little imprecise, but yes. As x gets infinitely large, k/x will ALWAYS approach zero.



In the case of f(x) = 17/x, it will still approach zero. 17/9999999999999999 is a very small positive number that's slightly above the x-axis.



In the case of f(x) = -17/x, it will also approach zero. The kicker is that -17/9999999999 is a very small NEGATIVE number that's slightly BELOW the x-axis.



Thus the first case, the graph approaches zero from above. One can say that lim x->inf (17/x) = 0+, where the + is tiny and almost a superscript. One can also say that lim x-->inf (-17/x) = 0-, where the - is a tiny superscript too.



However, don't say 17/inf = 0, or -17/inf = 0. It's an abuse of notation. Infinity's not a real number. Division by infinity isn't defined, since for division you need two numbers (and the second number CAN'T be zero!) Using the notation of limits is a much more accurate way to express these ideas.

Answer 2

What you have right that's a decrease at infinity of a rational function. In cases like those, you seem at 2 issues: a million) the utmost means of x interior the numerator 2) the utmost means of x interior the denominator for this reason, the utmost means of x in the two the numerator and denominator is two. Then the rule for this manner of question says you divide the coefficient of x^2 interior the numerator via the coefficient of x^2 interior the denominator. this would be a million/a million = a million. So, your decrease at infinity is a million. At -infinity, you are able to would desire to observe your signs and warning signs, yet for this reason you do no longer because of the fact the utmost means is two on the suitable and backside, so easily you're squaring issues, which makes each thing beneficial (or a minimum of nonnegative). Then your decrease at -infinity is likewise a million. The decrease is a million in the two case. wish this enables. EDIT: If the utmost means of x interior the numerator became smaller than the utmost means of x interior the denominator, then your decrease is 0. If the utmost means of x interior the numerator became larger than the utmost means of x interior the denominator, then your decrease is infinity (or -infinity counting on signs and warning signs and such..) those are the guidelines in the different 2 cases.

Answer 3

Yes, the limit (you must use the term limit because nothing can be divided by infinity) of any integer or expression divided by n as n --> infinite is 0. Think about it this way: as you increase the denominator, the answer decreases. As the denominator approaches infinity, the limit approaches 0.

Answer 4

You can't really divide by infinity, because infinity is not a real number.



We talk about the limit of 1/n, as n --> ∞, is 0, but we are not really dividing by ∞.



As long as the numerator a is finite, lim a/n --> 0 as n --> ∞.

Answer 5

17 = 17*1



Assuming 1/∞ = 0

1/∞ = 0

(17*1)/∞ = 17*0

17/∞ = 0



Same goes for any constant.



1/∞ = 0

(C*1)/∞ = C*0

C/∞ = 0



Also, 1/0 is undefined, not infinite. There's a difference.

If 1/0 = ∞

Then 1 = 0*∞

And therefore, 0 = 1

Doesn't hold ground. Therefore, the proof that....

If 1/0 = ∞

Then 1/∞ = 0

Is invalid.

Answer 6

Yeh, I'd agree with that. Basically, you are asking for the answer to the following:



lim (a/x) as x → ∞, where a is some constant.



The answer to the above is 0 (zero).

Answer 7

yes, any finite value divided by infinity yields 0.



17/∞ = 17*(1/∞) = 17*0 = 0



a/∞ = a*(1/∞) = a*0 = 0