Finding the limit without L'Hôpital's rule

[Unredeemed]

Homework Statement

Required to prove that
$\displaystyle\lim_{n\rightarrow \infty} ((1 - \frac{1}{n^2})^{n}) = 1$

Homework Equations

$\displaystyle\lim_{n\rightarrow \infty} ((1 + \frac{1}{n})^{n})$ is bounded above by e. I'm not sure if this is relevant, but it was the first part of the question, so I'd assume so?

Also, we haven't proved L'Hopital's rule yet, so I can't use that.

The Attempt at a Solution

I was thinking to maybe try and write it in a similar way to the first part.

So: $\displaystyle\lim_{n\rightarrow \infty} (((1 + \frac{1}{-(n^2)})^{-(n^2)})^{\frac{-1}{n}})$

But, as n tends to infinity $-n^2$ tends to negative infinity?

 

[Curious3141]

Homework Statement

Required to prove that
$\displaystyle\lim_{n\rightarrow \infty} ((1 - \frac{1}{n^2})^{n}) = 1$

Homework Equations

$\displaystyle\lim_{n\rightarrow \infty} ((1 + \frac{1}{n})^{n})$ is bounded above by e. I'm not sure if this is relevant, but it was the first part of the question, so I'd assume so?

Also, we haven't proved L'Hopital's rule yet, so I can't use that.

The Attempt at a Solution

I was thinking to maybe try and write it in a similar way to the first part.

So: $\displaystyle\lim_{n\rightarrow \infty} (((1 + \frac{1}{-(n^2)})^{-(n^2)})^{\frac{-1}{n}})$

But, as n tends to infinity $-n^2$ tends to negative infinity?

Try the generalised binomial theorem. If you want to prove rigorously that all the terms except the first go to zero at the limit, use the squeeze theorem.

 

[thchian]

$\displaystyle\lim_{n\rightarrow \infty} ((1 + \frac{1}{n})^{n})$ is bounded above by e. I'm not sure if this is relevant, but it was the first part of the question, so I'd assume so?

yes, it is relevant...firstly, u should understand why $\displaystyle\lim_{n\rightarrow \infty} ((1 + \frac{1}{n})^{n}) = e$ then u can solve this question easily..