Infinity Limit

[Oceanamie]

Could someone help me solve this?

lim as n -> infinity of [ a^(n+1) + b^(n+1) ] / [ (a^n) + (b^n) ]

 

[mmm4444bot]

Hello Oceanamie:

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[Oceanamie]

Hello Oceanamie:

This is a tutoring web site; we request that you follow the forum guidelines. Please read this summary page, and check out the links on that page for the complete guidelines and rules.

Then show us what you've tried so far, or explain why you're stuck.

Thank you! :cool:

I currently simplified to:

lim as n -> infinity of b * (1 + (a^(n+1)/b^(n+1) ) / (1 + (a^(n)/b^(n) )

So when you take the limit, that makes it just b. I am wondering if that's the correct answer, or not.

 

[stapel]

I currently simplified to:

lim as n -> infinity of b * (1 + (a^(n+1)/b^(n+1) ) / (1 + (a^(n)/b^(n) )

So when you take the limit, that makes it just b. I am wondering if that's the correct answer, or not.

If you think about it, you could have divided everything by \(\displaystyle a^{n+1}\) and ended up with "just a". If the method of simplification changes the answer, then probably there's something wrong with the method or the answer.

I don't disagree with the division technique, but on what basis have you decided that the limit will be "just b"? Are there some conditions on "a" and "b"?

 

[pka]

I currently simplified to:

lim as n -> infinity of b * (1 + (a^(n+1)/b^(n+1) ) / (1 + (a^(n)/b^(n) )

So when you take the limit, that makes it just b. I am wondering if that's the correct answer, or not.

It is not that simple. This is a concept question. You really need to consider several cases.

For example: \(\displaystyle a > b > 0\). What if \(\displaystyle a > 0> b ~?\)
ETC.

 

[Oceanamie]

I'm sorry; the conditions are: 0 < a < b

 

[pka]

I'm sorry; the conditions are: 0 < a < b

\(\displaystyle \dfrac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}} = \dfrac{{a{{\left( {\dfrac{a}{b}} \right)}^n} + b}}{{{{\left( {\dfrac{a}{b}} \right)}^n} + 1}}\)