anonymous
  • anonymous
I have two more limit problems, and I should be done for the night. Would you mind helping me?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
I even have the answer to this problem. I just don't understand the steps on how to solve it.
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zzr0ck3r
  • zzr0ck3r
which step do you not understand?

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anonymous
  • anonymous
Steps 1 and 2.
zzr0ck3r
  • zzr0ck3r
\(\dfrac{\frac{1}{a+b}-\frac{1}{a}}{b}=\dfrac{\frac{a}{a}(\frac{1}{a+b})-(\frac{a+b}{a+b})\frac{1}{a}}{b}=\dfrac{\frac{a-(a+b)}{a(a+b)}}{b}=\dfrac{-b}{a(a+b)}*\dfrac{1}{b}=\dfrac{-1}{a(a+b)}\)
zzr0ck3r
  • zzr0ck3r
\(\dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\=\dfrac{-1}{x(x+\Delta x)}\)
zzr0ck3r
  • zzr0ck3r
\[\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}\]
anonymous
  • anonymous
Haha yeah just trying to collect and process my thoughts.
zzr0ck3r
  • zzr0ck3r
lol :)
anonymous
  • anonymous
Could you explain how you got from \[\frac{ \frac{ 1 }{ x+x \Delta } -\frac{ 1 }{ x }}{ x \Delta }\] to the part after the = sign?
anonymous
  • anonymous
(sorry I would type it out, but it may take all night)
anonymous
  • anonymous
From the first circle to the second circle.
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anonymous
  • anonymous
Well, it's 3am where I'm at, so I think I'll just leave this topic open for now and check it again in the morning.
anonymous
  • anonymous
You get from the first circle to thee second by multiplying and dividing a fraction for the same quantity, you see? a/a and (a+b)/(a+b), if you multiply and divide for the same quantity (just like adding and subtracting the same quantity) the result doesn't change, but it often becomes easier to find!
zzr0ck3r
  • zzr0ck3r
\(\large \dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}\) Ok I want to get a common denominator which will be \(x(x+\Delta x)\). On the left hand side of the numerator I need to multiply top and bottom by \(x\) to make that happen. On the right ahnd side of the numerator I need to multiply top and bottom by \((x+\Delta x)\). \(\large =\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\ \\ \large =\dfrac{-1}{x(x+\Delta x)}\)
thomas5267
  • thomas5267
The step is trying to bring the two fractions together. \[ \frac{1}{3}-\frac{1}{5}=\frac{1}{3}\frac{5}{5}-\frac{1}{5}\frac{3}{3}=\frac{5-3}{15}=\frac{2}{15}\\ \frac{1}{x+\Delta x}-\frac{1}{x}=\frac{1}{x+\Delta x}\frac{x}{x}-\frac{1}{x}\frac{x+\Delta x}{x+ \Delta x}=\frac{-\Delta x}{x(x+\Delta x)}\\ \frac{x}{x}=\frac{x+\Delta x}{x+\Delta x}=1 \]
zzr0ck3r
  • zzr0ck3r
Right^^^
zzr0ck3r
  • zzr0ck3r
This is why i wrote \(\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-cb}{bd}\) ...Same thing :)
thomas5267
  • thomas5267
I really wish the OP has a temporary brain short circuit or else he/she will be having a horrible time dealing with calculus.
zzr0ck3r
  • zzr0ck3r
The more you do it, the more you get used to it... Calculus helps solidify algebra skills because so much of it rests on long tedious algebra.
zzr0ck3r
  • zzr0ck3r
Like day one and the formal definition of the derivative....
zzr0ck3r
  • zzr0ck3r
somewhat formal...
thomas5267
  • thomas5267
That looks like the left hand limit of the derivative of \(\dfrac{1}{x}\) if you get what I mean.
anonymous
  • anonymous
@zzr0ck3r Common denominator! Thank you. For some reason (probably lack of sleep) I couldn't tell what was going on. I feel stupid for not being able to tell. Sorry about that. @thomas5267 Yes, it was a temporary brain short circuit. As I stated earlier, it was 3 in the morning when I was trying to do this, so something so simple just easily slipped my mind.

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