Proof of $\lim_{k\to \infty}{|x^k-x|\over { 1+|x^k-x|}}=0$

Question

Is there an easy way to prove that $$\lim_{k\to \infty}{|x^k-x|\over { 1+|x^k-x|}}=0$$ with $x\in \mathbb R$ without using $\epsilon-\delta$ definition? Any help would be really appreciated

Answer 1

If $|x|<1$, then $|x^k-x|=x-x^k$ $$\lim_{k\rightarrow\infty}\frac{x-x^k}{1+x-x^k}=\frac{x}{1+x}.$$ If $|x|=1$, then $$\lim_{k\rightarrow\infty}\frac{|x^k-x|}{1+|x^k-x|}=0.$$ If $|x|>1$, then $|x^k-x|=x^k-x$ $$\lim_{k\rightarrow\infty}\frac{x^k-x}{1+x^k-x}=1-\lim_{k\rightarrow\infty}\frac{1}{1+x^k-x}=1.$$

Answer 2

If $x>1$ or $0 > x > -1$ and $k \to \infty$, $|x^k-x|=x^k-x$ since the thing inside the absolute value will be positive.

If $x<-1$ or $0 < x<1$ and $k \to \infty$, $|x^k-x|=x-x^k$ since the thing inside the absolute value will be negative.

If $x=0$ or $x=1$, $|x^k-x|=0$

Do two different limits, one for each case, and you will find your answer.

Thanks to Moya for pointing out the error.

Answer 3

If $x > 1$, then this is false, because ${|x^k-x|\over { 1+|x^k-x|}} ={-1+1+|x^k-x|\over { 1+|x^k-x|}} ={-1\over { 1+|x^k-x|}}+1 $ and ${-1\over { 1+|x^k-x|}} \to 0 $ as $k \to \infty $ since $|x^k-x| \to \infty $.