Another find the limit

[quasar987]

I've tried multiplicating by the conjugate of the denominator and of the numerator but this leads to nothing I can see. How can this limit be evaluated? (the limit is to be taken when n goes to infinity)

\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}

The answer is 1.

Thanks for helping.

[Hurkyl]

I've tried multiplicating by the conjugate of the denominator and of the numerator but this leads to nothing I can see.

Show us what you got when you did this.

[Leong]

Divide the numerator and the denominator by \sqrt{n}.

[quasar987]

Show us what you got when you did this.
Will do tomorrow. I gotta go to bed urgent.

Divide the numerator and the denominator by \sqrt{n}.
If you mean "take \sqrt{n} out of the num and denom", when you take the limit you get the undeterminate form 0/0. If that's not what you mean, I don't know what you mean. :smile:

[quasar987]

Ok, so

\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}} = \frac{\sqrt{n+7}\sqrt{n+2}+\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n+2}-\sqrt{n+5}\sqrt{n}}{2}

which may be factorised into...

\frac{(\sqrt{n+7}-\sqrt{n+5})(\sqrt{n+2}+\sqrt{n})}{2}


If we multiply by the other conjugate, we get

\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}} \frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+\sqrt{n+5} } = \frac{12}{\sqrt{n+2}\sqrt{n+7}+\sqrt{n+2}\sqrt{n+5 } -\sqrt{n+7}\sqrt{n}-\sqrt{n+5}\sqrt{n}}

which may be factorised into

\frac{12}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+5}+\sqrt{n+7})}

[arildno]

You've made quite a few errors here
Multiply with 1*1 in this manner:
\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+7}+\sqrt{n+5}}{\sqrt{n+7}+ \sqrt{n+5}}*\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}*\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\s qrt{n}}
Hence, we get:
\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+7}+\s qrt{n+5}}

[quasar987]

And THEN...
Divide the numerator and the denominator by \sqrt{n}. :tongue:

Ok I get it now. Thanks everyone!