- Thread starter Axel
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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

Given: Y = (ax + 2)/x

Find: limit [x → ∞] Y

Find: limit [x → 0] Y

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limit [x → ∞] Y = a \(\quad\) Correctx to infinity y=a ?

x to 0 y=2 ?

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

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

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

x to 0 y=2 ?

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

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

-Dan

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.

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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 denominatorB. 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 toBut 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.

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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.limit [x → 0] Y = 2/x

-Dan