What is the limit of the given expression when n goes to infinity?

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Homework Help Overview

The discussion revolves around evaluating the limit of a given expression as n approaches infinity, specifically the limit of the expression \(\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}\).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for simplifying the expression, including multiplying by conjugates and dividing by \(\sqrt{n}\). Some express confusion over the results of their attempts, particularly regarding the indeterminate form encountered.

Discussion Status

There is ongoing exploration of different algebraic manipulations to evaluate the limit. Some participants have provided guidance on specific steps, while others are questioning the validity of certain approaches. The discussion reflects a collaborative effort to clarify the problem without reaching a definitive conclusion.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of assistance they can provide to one another.

quasar987
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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)

[tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}[/tex]

The answer is 1.

Thanks for helping.
 
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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.
 
Divide the numerator and the denominator by [tex]\sqrt{n}[/tex].
 
Hurkyl said:
Show us what you got when you did this.
Will do tomorrow. I got to go to bed urgent.

Leong said:
Divide the numerator and the denominator by [tex]\sqrt{n}[/tex].
If you mean "take [itex]\sqrt{n}[/itex] 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:
 
Ok, so

[tex]\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}[/tex]

which may be factorised into...

[tex]\frac{(\sqrt{n+7}-\sqrt{n+5})(\sqrt{n+2}+\sqrt{n})}{2}[/tex]


If we multiply by the other conjugate, we get

[tex]\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}}[/tex]

which may be factorised into

[tex]\frac{12}{(\sqrt{n+2}-\sqrt{n})(\sqrt{n+5}+\sqrt{n+7})}[/tex]
 
You've made quite a few errors here
Multiply with 1*1 in this manner:
[tex]\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}+\sqrt{n}}[/tex]
Hence, we get:
[tex]\frac{\sqrt{n+7}-\sqrt{n+5}}{\sqrt{n+2}-\sqrt{n}}=\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+7}+\sqrt{n+5}}[/tex]
 
Last edited:
And THEN...
Leong said:
Divide the numerator and the denominator by [tex]\sqrt{n}[/tex].
:-p

Ok I get it now. Thanks everyone!
 

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