I'm trying to solve the following exercise from Tao's Analysis 1 textbook:
"Let $x$ be a real number. Then $\lim_{n\to\infty} x^n$ exists and is equal to zero when $|x|<1$, exists and is equal to $1$ when $x=1$, and diverges when $x=-1$ or when
$|x|>1$."
DEF.(Limits of sequences) If a sequence $(a_n)_{n=m}^\infty$ converges to some real number $L$ we say that $(a_n)_{n=m}^\infty$ is convergent and that its limit is $L$; we write $\lim_{n\to\infty}a_n=L$. If a sequence $(a_n)_{n=m}^\infty$ is not converging to any real number $L$, we say that the sequence $(a_n)_{n=m}^\infty$ is divergent and we leave $\lim_{n\to\infty} a_n$ undefined.
I've been able to prove the statement for $|x|<1$, $x=1$, $x=-1$, $x>1$, but I haven't been able to do the same for the case $x<-1$, so I would appreciate any hint about how to do this last case.
(NOTE: the concept of subsequence is introduced later in the textbook, so it cannot be used to solve this exercise)
Best regards,
lorenzo.
Best Answer
Let's assume for contradiction that $\lim_{n \to \infty}(-x)^n$ converges to $l$ for $x>1$, then $\lim_{n \to \infty}|(-x)^n| = |l|$ (I hope you can use this). Finally, $|(-x)^n| = |x^n|$ $\implies$ $\lim_{n \to \infty}x^n = |l|$, which is a contradiction, because $\lim_{n \to \infty}x^n$ does not converge when $x>1$ (as you has proved before).