# Math Help - Limit

1. ## Limit

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2. What ideas have you had so far?

3. Originally Posted by Ackbeet
What ideas have you had so far?
I am trying to find the equation of the line that pass through (a/2 ,a^2/2) and (0,b)

I am trying to find the equation of the line that pass through (a/2 ,a^2/2) and (0,b)
Not a bad way to start, except that the line will go through $\left(\frac{a}{2},\frac{a^{2}}{4}\right)$ not $\left(\frac{a}{2},\frac{a^{2}}{2}\right).$ What do you get?

5. Originally Posted by Ackbeet
Not a bad way to start, except that the line will go through $\left(\frac{a}{2},\frac{a^{2}}{4}\right)$ not $\left(\frac{a}{2},\frac{a^{2}}{2}\right).$ What do you get?
i get y=(2a^2-8b/4a)x+b

i get y=(2a^2-8b/4a)x+b
I don't think you've quite got it yet. If you plug in $x=a/2,$ you don't get $y=a^{2}/4,$ like you should. Can you show your work?

7. Originally Posted by Ackbeet
Not a bad way to start, except that the line will go through $\left(\frac{a}{2},\frac{a^{2}}{4}\right)$ not $\left(\frac{a}{2},\frac{a^{2}}{2}\right).$ What do you get?
Actually it is $\left(\frac{a}{2},\frac{a^{2}}{2}\right).$.
Because we want the midpoint of the line segment between $(0,0),~\&~(a,a^2)$.

8. Originally Posted by Plato
Actually it is $\left(\frac{a}{2},\frac{a^{2}}{2}\right).$.
Because we want the midpoint of the line segment between $(0,0),~\&~(a,a^2)$.
Ach! You're right. The midpoint should be on the line, not the function. BAHADEEN's line is still incorrect, though, because I get $a^{3}$ for y when I plug in $x=a/2$.

Originally Posted by topsquark
Ack! Beet beat me again!
How about "Ack! Beet-red beat me again!"? Like Deveno, you could even put that in your signature if you wanted.