# y = (ax+2) / x x to 0 Solution anyone?

#### Axel

##### New member
Y=(ax+2)/x
when x to infinity
and x go to 0

#### Axel

##### New member
x to infinity y=a ?
x to 0 y=2 ?

#### Otis

##### Senior Member
Y=(ax+2)/x
when x to infinity
and x go to 0
Hello Axel. You ought to have typed the word 'limit' somewhere, yes?

Given: Y = (ax + 2)/x

Find: limit [x → ∞] Y

Find: limit [x → 0] Y #### Otis

##### Senior Member
x to infinity y=a ?
x to 0 y=2 ?
limit [x → ∞] Y = a $$\quad$$ Correct

limit [x → 0] Y = 2 $$\quad$$ Incorrect

Please show your work, so that we may see how you arrived at 2. #### topsquark

##### Full Member
x to infinity y=a ?
x to 0 y=2 ?
The first part is correct. For limit as x goes to 0...

$$\displaystyle \lim_{x \to 0} \dfrac{ax + 2}{x} = \lim_{x \to 0} \left ( a + \dfrac{2}{x} \right )$$

What is the limit of the second term as x goes to 0?

-Dan

#### Axel

##### New member
A. When x goes to infinity. The 2 x's take out each other. And I remove them. Add 2 to infinity won't give anything more. a is left.

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.
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.

#### HallsofIvy

##### Elite Member
A. When x goes to infinity. The 2 x's take out each other.And I remove them.
I cringe at this wording! I think what you mean when you say "the 2 x's take each other out" is that $$\displaystyle \frac{ax+ 2}{x}= \frac{ax}{x}+ \frac{2}{x}= a+ \frac{2}{x}$$.

Add 2 to infinity won't give anything more. a is left.
As x goes to infinity, 2/x goes to 0.

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.
x "so small" in the denominator cannot "be excluded"!

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.
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.

• topsquark

#### Axel

##### New member
Given: Y = (ax + 2)/x
Find: limit [x → ∞] Y
Find: limit [x → 0] Y

limit [x → ∞] Y = a
limit [x → 0] Y = 2/x

I admire your capacity. I went in to brain meltdown. A BIG thank you and a nice weekend. //Axel

#### topsquark

##### Full Member
limit [x → 0] Y = 2/x
No! $$\displaystyle \lim_{x \to 0} \dfrac{2}{x}$$ 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.

-Dan