Help with limit approaching infinity problem

bananope

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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
 
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 xx that figurates in fraction.
 
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 xx+12x\dfrac{ \sqrt{x} }{ \sqrt{x + 1} } - \sqrt{2x}. I suspect you want
(sqrt(x))/(sqrt(x + 1) + sqrt(2x)), which is xx+1+2x\dfrac{ \sqrt{x} }{ \sqrt{x + 1} + \sqrt{2x} }.

So.
xx+1+2x12x12x\dfrac{ \sqrt{x} }{ \sqrt{x + 1} + \sqrt{2x} } \cdot \dfrac{ \dfrac{1}{ \sqrt{2x} } }{ \dfrac{1}{ \sqrt{2x} } }

=x2xx+12x+1= \dfrac{ \sqrt{ \dfrac{x}{2x} } }{ \sqrt{ \dfrac{x + 1}{2x} } + 1 }

Now, x+12x=x2x+12x=12+12x\dfrac{x + 1}{2x} = \dfrac{x}{2x} + \dfrac{1}{2x} = \dfrac{1}{2} + \dfrac{1}{2x}

=1212+12x+1= \dfrac{ \sqrt{ \dfrac{1}{2} } }{ \sqrt{ \dfrac{1}{2} + \dfrac{1}{2x} } + 1 }

What happens to this as x goes to infinity?

-Dan
 
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