Help with limit approaching infinity problem

[bananope]

limit of x approaches infinity (sqrt(x))/sqrt(x+1)-sqrt(2x)

So far, I multiplied both sides by 1/sqrt(2x) to get sqrt(1/2)/sqrt(x+1/2x)-1

I don't know how to simplify further to find the limit

 

[Pedja]

You should find common denominator for these two terms and rewrite expression into one fraction, and then divide both the numerator and denominator by the largest power of [imath]x[/imath] that figurates in fraction.

 

[skeeter]

Try dividing each term in the numerator and denominator by [imath]\sqrt{x}[/imath], then take the limit ...

 

[topsquark]

limit of x approaches infinity (sqrt(x))/sqrt(x+1)-sqrt(2x)

So far, I multiplied both sides by 1/sqrt(2x) to get sqrt(1/2)/sqrt(x+1/2x)-1

I don't know how to simplify further to find the limit

I suspect your problem is written wrong. You wrote (sqrt(x))/sqrt(x + 1) - sqrt(2x) which is [imath]\dfrac{ \sqrt{x} }{ \sqrt{x + 1} } - \sqrt{2x}[/imath]. I suspect you want
(sqrt(x))/(sqrt(x + 1) + sqrt(2x)), which is [imath]\dfrac{ \sqrt{x} }{ \sqrt{x + 1} + \sqrt{2x} }[/imath].

So.
[imath]\dfrac{ \sqrt{x} }{ \sqrt{x + 1} + \sqrt{2x} } \cdot \dfrac{ \dfrac{1}{ \sqrt{2x} } }{ \dfrac{1}{ \sqrt{2x} } }[/imath]

[imath]= \dfrac{ \sqrt{ \dfrac{x}{2x} } }{ \sqrt{ \dfrac{x + 1}{2x} } + 1 }[/imath]

Now, [imath]\dfrac{x + 1}{2x} = \dfrac{x}{2x} + \dfrac{1}{2x} = \dfrac{1}{2} + \dfrac{1}{2x}[/imath]

[imath]= \dfrac{ \sqrt{ \dfrac{1}{2} } }{ \sqrt{ \dfrac{1}{2} + \dfrac{1}{2x} } + 1 }[/imath]

What happens to this as x goes to infinity?

-Dan