anonymous
  • anonymous
lim(y->0), (sqrt(x+y)-sqrt(x))/y Basically, I'm trying to find the integral of sqrt(x) without using actual calculus.... which I personally don't get, but whatever. Any ideas??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
You're just going to have to muck around with algebra.
anonymous
  • anonymous
try by multiplying the conjugate.
anonymous
  • anonymous
*bangs head on wall* dammit...... I tried but once y approaches zero then it starts to become dividing by zero... and then people start catching on fire.

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anonymous
  • anonymous
No they don't. It's a limit. You're not actually dividing by zero.
anonymous
  • anonymous
Well, true. DNE but still i need a function for an answer
anonymous
  • anonymous
I dont think that limit will give integral of sqrt(x)
anonymous
  • anonymous
If you want to find derivative then its {sqrt(x+y)-sqrt(x)}/y not {sqrt(x+y)+sqrt(x)}/y as y approaches 0
anonymous
  • anonymous
ah, did I type it wrong? my bad... you're right, its - not +
anonymous
  • anonymous
\[ \begin{split} \lim_{y\to0}\frac{\sqrt{x+y}-\sqrt{x}}y &= \lim_{y\to0}\frac{(\sqrt{x+y}-\sqrt{x})(\sqrt{x+y}+\sqrt{x})}{y(\sqrt{x+y}+\sqrt{x})}\\ &= \lim_{y\to0}\frac{x+y-x}{y(\sqrt{x+y}+\sqrt{x})}\\ &=\lim_{y\to0}\frac{y}{y(\sqrt{x+y}+\sqrt{x})}\\ &=\lim_{y\to0}\frac{1}{\sqrt{x+y}+\sqrt{x}}\\ &=\frac{1}{\sqrt{x}+\sqrt{x}}\\ &=\frac{1}{2\sqrt{x}} \end{split} \]
anonymous
  • anonymous
thats correct
anonymous
  • anonymous
oooooh....... brilliant brilliant!
anonymous
  • anonymous
the second step numerator is what I had wrong...

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