Question

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??

Answer 1

You're just going to have to muck around with algebra.

Answer 2

try by multiplying the conjugate.

Answer 3

*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.

Answer 4

No they don't. It's a limit. You're not actually dividing by zero.

Answer 5

Well, true. DNE but still i need a function for an answer

Answer 6

I dont think that limit will give integral of sqrt(x)

Answer 7

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

Answer 8

ah, did I type it wrong? my bad... you're right, its - not +

Answer 9

\[
\begin{split}
\lim_{y\to0}\frac{\sqrt{x+y}-\sqrt{x}}y
&= \lim_{y\to0}\frac{(\sqrt{x+y}-\sqrt{x})(\sqrt{x+y}+\sqrt{x})}{y(\sqrt{x+y}+\sqrt{x})}\\
&= \lim_{y\to0}\frac{x+y-x}{y(\sqrt{x+y}+\sqrt{x})}\\
&=\lim_{y\to0}\frac{y}{y(\sqrt{x+y}+\sqrt{x})}\\
&=\lim_{y\to0}\frac{1}{\sqrt{x+y}+\sqrt{x}}\\
&=\frac{1}{\sqrt{x}+\sqrt{x}}\\
&=\frac{1}{2\sqrt{x}}
\end{split}
\]

Answer 10

thats correct

Answer 11

oooooh....... brilliant brilliant!

Answer 12

the second step numerator is what I had wrong...