Limit Epsilon Delta

Question

everybody!
I am a new user here. Please correct me if I make any mistakes.

Show that for any $\epsilon$>0 there exists N such that for all n $\geqq$ N it is true that $|x^n - 0|$<$\epsilon$

x $\in$ (-1,1)

$x^n \to$ 0 as n$\to$ ∞

I tried to solve this problem
$lim_{x\to ∞}$ $x^n$ = 0
|$x^n$-0|<$\epsilon$
$x^n$ < $\epsilon$
But, I am not sure how to continue the proof.

I also have another question:
$x_n$ = $\frac {a^n - b^n}{a - b}$
$(\frac{b}{a})^n$ $\to$ 0 as n $\to$ ∞

Show that for any integer k $\geqq$ 1
$\frac {x_n+_k}{x_n}$ $\to$ $a^k$ as n $\to$ ∞

This is what I did:
And I got $lim_{n\to ∞}$ $\frac {a^n{^+}^k - b^n{^+}^k}{a^n - b^n}$

Again, I am stuck as I don't know how to finish it. Can someone please direct me step-by-step? I need to understand this topic well. Thank you very much.

Answer 1

For the second part note that $a^{n}-b^{n}$ has a factor of $a-b$. You can find the factorisation using the binomial theorem. Then taking the limit becomes trivial.

Answer 2

For the first problem, let $\epsilon>0$ then pick $n>\dfrac{\ln\epsilon}{\ln|x|}$. Then, since for $|x|<1$, $\ln|x|<0$ it follows that $n\ln|x|<\ln\epsilon$. So $\ln|x|^n<\ln\epsilon$. Since $y=e^x$ is an increasing function $\ln|x|^n<\ln\epsilon$ if and only if $e^{\ln|x|^n}<e^{\ln\epsilon}$. That is to say, if and only if $|x|^n<\epsilon$

Answer 3

For the second one you can do this.

$x_{n+k}=\frac{a^{n+k}-b^{n+k}}{a-b}$

So the given expression is

$\frac{x_{n+k}}{x_n}=\frac{a^{n+k}-b^{n+k}}{a^n-b^n}$

Taking $a^{n+k}$ common in numerator and $a^n$ in denominator we get

$\lim: \lim_{n\to \infty} \frac{a^k(1-\left({\frac{b}{a}}\right)^{n+k})}{1-\left({\frac{b}{a}}\right)^n}$

Since $\frac{b}{a}\to0$ as $n\to\infty$

The $\frac{b}{a}$ terms are 0 and we get the limit $a^k$.