Sequence Limit Problem

[Jason76]

\(\displaystyle a_{n} = 2 + (0.1)^{n}\)

Converge or Diverge, and if converge, then to what value?

\(\displaystyle \lim n \rightarrow \infty[ (2 + (0.1)^{n}]\)

\(\displaystyle \lim n \rightarrow \infty[(2 + (0.1)^{\infty}] = \infty\)

From immediately plugging in infinity, it seems to diverge, but the book answer is 2.

 

[stapel]

\(\displaystyle a_{n} = 2 + (0.1)^{n}\)

Converge or Diverge, and if converge, then to what value?

Since 2 is constant, you only really need to figure out what happens to (1/10)^n.

\(\displaystyle \lim n \rightarrow \infty[ (2 + (0.1)^{n}]\)

\(\displaystyle \lim n \rightarrow \infty[(2 + (0.1)^{\infty}] = \infty\)

From immediately plugging in infinity, it seems to diverge, but the book answer is 2.

How did you get that (1/10)^n gets infinitely large as n increases? What did you get when you graphed f(x) = (1/10)^x?

 

[Jason76]

Since 2 is constant, you only really need to figure out what happens to (1/10)^n.


How did you get that (1/10)^n gets infinitely large as n increases? What did you get when you graphed f(x) = (1/10)^x?

Would one rewrite \(\displaystyle (.1)^{\infty}\) as \(\displaystyle 1/10^{\infty}\)? In that case, the \(\displaystyle 1/10\) stuff would evaluate to 0, thus leaving 2, the constant, as an answer.

\(\displaystyle \dfrac{1^{n}}{10^{n}} = (1/10)^{n}\)

\(\displaystyle \dfrac{1^{\infty}}{10^{\infty}} \) indeterminate

Maybe you could prove this idea by using l hopital's, hence, \(\displaystyle \dfrac{d}{dx}(10^{n}) = 10^{n} \ln 10 (1)\) and \(\displaystyle 1\), in the numerator, would have a derivative of 0, hence the limit comes out to 0.

But I think you would need L hopitals again cause, you the limit would come out to \(\displaystyle \dfrac{0}{\infty}\) which is indeterminate again.

If you did the product rule for the denominator (on the 2nd go around of L hopitals), maybe it comes out to some actual number, then an actual number divided by 0 (the numerator would stay 0) would be 0.

 

[HallsofIvy]

Please stop writing \(\displaystyle \infty\) as if it were actually a number! It isn't and you cannot do any "arithmetic" with it such as raising to powers and making fractions. These are all limit problems, not "evaluating at \(\displaystyle \infty\)".

 

[Jason76]

Please stop writing \(\displaystyle \infty\) as if it were actually a number! It isn't and you cannot do any "arithmetic" with it such as raising to powers and making fractions. These are all limit problems, not "evaluating at \(\displaystyle \infty\)".

How are we to think about \(\displaystyle \infty\), in terms of this limit problem?

So the best way to think, is that the limit as n approaches infinity of [\(\displaystyle 1/\)some number] , is always going toward 0.

 

[stapel]

How are we to think about \(\displaystyle \infty\), in terms of this limit problem?

The value of [whatever in terms of n] as n "gets arbitrarily large".

 

[HallsofIvy]

Surely you could have seen that 2+ .1= 2.1, 2+ .01= 2.01, 2+ .001= 2.001, 2+ .0001= 2.0001, etc. It's not hard to see what that's getting closer and closer to.