Limit, as n tends to infinity, of n - sqrt (n^2 - n)

[ku1005]

Hey guys,

with the following limit im just having abit of trouble and was hoping for some clarification, since I feel i am missing something silly

Calculate the limit as n -> inifnity for

n - sqrt (n^2 - n).

I did the following:

= n - sqrt ((n^2)[1 - 1/n])

= n - |n|sqrt(1 - 1/n)

which thus suggests as n -> infinity that it becomes
inifity - infinity

which doesnt exist, however upon inspection with software, this limit seems to be 0.5. Any tips would be great!

cheers

 

[galactus]

\(\displaystyle \L\\\lim_{n\to\infty}[n-\sqrt{n^{2}-n}]\)

Multiply top and bottom by conjugate:

\(\displaystyle \L\\\frac{n-\sqrt{n^{2}-n}}{1}\cdot\frac{n+\sqrt{n^{2}-n}}{n+\sqrt{n^{2}-n}}\)

\(\displaystyle \L\\\frac{n}{n+\sqrt{n^{2}-n}}\)

Divide by n and \(\displaystyle \sqrt{n^{2}}\) and get:

\(\displaystyle \L\\\frac{\frac{n}{n}}{\frac{n}{n}+\frac{\sqrt{n^{2}-n}}{\sqrt{n^{2}}}}=\frac{1}{1+\sqrt{1-\frac{1}{n}}}\)

\(\displaystyle \L\\\lim_{n\to\infty}\frac{1}{1+\sqrt{1-\frac{1}{n}}}=\frac{1}{2}\)

 

[ku1005]

I see.....thanks for that!!