Math manipulation

1. Nov 14, 2011

sparkle123

Why does

when l approaches infinity?
This would mean that but I don't understand why that would be true.
Thanks!!

2. Nov 14, 2011

Dick

sqrt(1+r)=1+r/2-r^2/8+... That's the Taylor series expansion of sqrt(1+r). If r is small, then 1+r/2 is a good approximation. And as l->infinity then y^2/(l-x)^2 becomes very small. The sides aren't equal for any finite l. But the approximation becomes better and better as l->infinity.

3. Nov 14, 2011

ehild

Try it: Calculate the square root of 1+x with smaller and smaller x-values. For example, √1.2=1.095, √1.02=1.0099, √1.002=1.000999....

$\sqrt{1+x} \approx 1+x/2$

ehild

4. Nov 15, 2011

sparkle123

Thank you Dick and ehild! :)