Question

Pointwise convergence

Answer 1

Can someone help my find the limit of this:
\[\sqrt{n^2+nx}-n\]

Answer 2

Interesting :D
\[\Large = \sqrt{n^2\left(1+\frac{x}n\right)}-n\]

Answer 3

\[\Large n\sqrt{1+\frac{x}n}-n\]

Answer 4

\[\Large = n\left(\sqrt{1+\frac{x}n}-1\right)\]

Answer 5

Better? :D

Answer 6

Hmm...

Answer 7

I know it convergence against x/2. But cant find a way to see it.

Answer 8

Is it? I don't really know, hang on.. :D

Answer 9

\[\Large \frac{\sqrt{1+\frac{x}n}-1}{\frac1n}\]

Answer 10

differentiating both the numerator and the denominator with respect to n...
\[\Large \frac{\frac1{2\sqrt{1+\frac{x}n}}\cdot\frac{-x}{n^2}}{-\frac1{n^2}}\]

Answer 11

If you simplify this...
\[\Large = \frac{x}{2\sqrt{1+\frac{x}n}}\]
And now take the limit as n goes to infinity XD

Answer 12

Thank you

Answer 13

@terenzreignz
Have do you come from this
\[\sqrt{n^2+nx}-n\]
to this
\[\Large \frac{\sqrt{1+\frac{x}n}-1}{\frac1n}\]