Hello Axel. You ought to have typed the word 'limit' somewhere, yes?Y=(ax+2)/x
when x to infinity
and x go to 0
limit [x → ∞] Y = a \(\quad\) Correctx to infinity y=a ?
x to 0 y=2 ?
The first part is correct. For limit as x goes to 0...x to infinity y=a ?
x to 0 y=2 ?
I cringe at this wording! I think what you mean when you say "the 2 x's take each other out" is that \(\displaystyle \frac{ax+ 2}{x}= \frac{ax}{x}+ \frac{2}{x}= a+ \frac{2}{x}\).A. When x goes to infinity. The 2 x's take out each other.And I remove them.
As x goes to infinity, 2/x goes to 0.Add 2 to infinity won't give anything more. a is left.
x "so small" in the denominator cannot "be excluded"!B. X goes to 0 Then thought ax in numerator will be so small and also x in denominator that ax and x could be excluded. 2 Is left.
You have, again, \(\displaystyle a+ \frac{2}{x}\). As x goes to 0, \(\displaystyle \frac{2}{x}\) goes to infinity. Adding a to that "wont give anything more". Depending upon which your teacher prefers, you can say that "the limit is infinity" or "the limit does not exist.But when I see topsquark I think 2/x will be very big so 2+(very big) ? Please more help. Thank you very much so far.
No! [math]\lim_{x \to 0} \dfrac{2}{x}[/math] is either "does not exist" or "is infinite." 2/x is not a possible answer because you still have an "x" in your proposed answer.limit [x → 0] Y = 2/x