Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider this limit: $$\lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n^2} = x$$

I thought the way to solve this for $x$ was to reduce it using the fact that as $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$: $$\therefore \lim_{n\rightarrow\infty}(1+0)^{n^2} = x$$

Apparently this is wrong! Why is it wrong?

share|improve this question
1  
By the same logic, $1=\lim_{n\rightarrow\infty}n\frac{1}{n}=\lim_{n\rightarrow\infty}n(0)=0$ which is clearly wrong. –  Daniel Rust May 13 '13 at 22:29
    
@DanielRust Clearly. Thanks! –  xisk May 13 '13 at 22:31
    
If it was $n\longrightarrow -\infty$ there would be a finite answer. –  Moron May 14 '13 at 10:03

3 Answers 3

up vote 3 down vote accepted

But as $n\rightarrow \infty $ that exponent gets ever larger. A smaller argument, yes, but raised to an ever larger power. Can't assume that the whole thing is going to go to one. Try a few examples on your calculator to see what happens.

For a little more detail, if $n$ is some big number, when you multiply everything out, the first two terms (look up the binomial expansion) will be

$1+ n^2 * \displaystyle\frac{1}{n} = 1+n \rightarrow \infty$

None of the remaining terms will be negative, so you know this is a bottom limit on what you are going to get.

share|improve this answer

If your method isn't wrong so we have also

$$\lim_{n\to\infty}\frac{1}{n}\times n=\lim_{n\to\infty}0\times n=0???$$

share|improve this answer

You have an indeterminate form: $1^\infty$, so need some finer analysis than just plugging in. We know $\displaystyle \lim_{n\rightarrow\infty} \left( 1+\frac{1}{n} \right) ^{n} = e$ so we might expect yours to diverge as $e^n$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.