Limes problem no. 1

[mathwannabe]

Hello everybody

I'm doing my course in Mathematical Analysis 1 (I guess that's Calculus 1 in western terminology) on my faculty. Currentlu we are doing limits of sequences and I stumbled upon a problem that I can't seem to solve. It goes like this:

1) \(\displaystyle \lim_{n \to \infty}\dfrac{1+a+...+a^n}{1+b+...+b^n}=?\)

So, I started by finding a formulas for sums of sequences in the numerator and in the donominator:

\(\displaystyle \lim_{n \to \infty}\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}\)

\(\displaystyle =\lim_{n \to \infty}\dfrac{(1-a^n)(1-b)}{(1-b^n)(1-a)}\)

\(\displaystyle =\lim_{n \to \infty}\dfrac{1-b-a^n+a^nb}{1-a-b^n+ab^n}\)

I get stuck here... I tried to factor \(\displaystyle b^n\) from both the numerator and the denominator, but I cant seem to get the solution provided by the book. Any hint?

Solution in the book: \(\displaystyle \dfrac{1-b}{1-a}\)

 

[girodd]

Hello everybody

I'm doing my course in Mathematical Analysis 1 (I guess that's Calculus 1 in western terminology) on my faculty. Currentlu we are doing limits of sequences and I stumbled upon a problem that I can't seem to solve. It goes like this:

1) \(\displaystyle \lim_{n \to \infty}\dfrac{1+a+...+a^n}{1+b+...+b^n}=?\)

So, I started by finding a formulas for sums of sequences in the numerator and in the donominator:

\(\displaystyle \lim_{n \to \infty}\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}\)

\(\displaystyle =\lim_{n \to \infty}\dfrac{(1-a^n)(1-b)}{(1-b^n)(1-a)}\)

\(\displaystyle =\lim_{n \to \infty}\dfrac{1-b-a^n+a^nb}{1-a-b^n+ab^n}\)

I get stuck here... I tried to factor \(\displaystyle b^n\) from both the numerator and the denominator, but I cant seem to get the solution provided by the book. Any hint?

Solution in the book: \(\displaystyle \dfrac{1-b}{1-a}\)

First of all, you need to consider cases. a^n -> 0 if |a| < 1, likewise for b. When this is the case it is simple to evaluate your expression. In all other cases, where |a| or |b| or both are >= 1, you need to analyze more carefully. In point of fact, for some of these cases there is no limit, and so the answer in the book cannot be the whole story.

 

[mathwannabe]

First of all, you need to consider cases. a^n -> 0 if |a| < 1, likewise for b. When this is the case it is simple to evaluate your expression. In all other cases, where |a| or |b| or both are >= 1, you need to analyze more carefully. In point of fact, for some of these cases there is no limit, and so the answer in the book cannot be the whole story.

Yes, I forgot to mention that abs(a), abs(b) < 1.

 

[HallsofIvy]

Hello everybody

I'm doing my course in Mathematical Analysis 1 (I guess that's Calculus 1 in western terminology) on my faculty. Currentlu we are doing limits of sequences and I stumbled upon a problem that I can't seem to solve. It goes like this:

1) \(\displaystyle \lim_{n \to \infty}\dfrac{1+a+...+a^n}{1+b+...+b^n}=?\)

So, I started by finding a formulas for sums of sequences in the numerator and in the donominator:

\(\displaystyle \lim_{n \to \infty}\dfrac{\dfrac{1-a^n}{1-a}}{\dfrac{1-b^n}{1-b}}\)

\(\displaystyle =\lim_{n \to \infty}\dfrac{(1-a^n)(1-b)}{(1-b^n)(1-a)}\)

Stop here! Now use the fact that, because all the functions involved are continuous, you can take the limit "inside" these functions- this limit is the same as
\(\displaystyle \dfrac{1- \lim_{n\to\infty}a^n}{1- a}\dfrac{1-b}{1- \lim_{n\to\infty}b^n}\)

and, of course, \(\displaystyle \lim_{n\to\infty}a^n= \lim_{n\to\infty}b^n= 0\).

\(\displaystyle =\lim_{n \to \infty}\dfrac{1-b-a^n+a^nb}{1-a-b^n+ab^n}\)

I get stuck here... I tried to factor \(\displaystyle b^n\) from both the numerator and the denominator, but I cant seem to get the solution provided by the book. Any hint?

Solution in the book: \(\displaystyle \dfrac{1-b}{1-a}\)

 

[mathwannabe]

Stop here! Now use the fact that, because all the functions involved are continuous, you can take the limit "inside" these functions- this limit is the same as
\(\displaystyle \dfrac{1- \lim_{n\to\infty}a^n}{1- a}\dfrac{1-b}{1- \lim_{n\to\infty}b^n}\)

and, of course, \(\displaystyle \lim_{n\to\infty}a^n= \lim_{n\to\infty}b^n= 0\).

Thank you, this helped me a lot.