Let $ \{ a_n \}_{n=m}^{\infty}$ be convergent sequence of real numbers such that $\lim _{n \rightarrow \infty}a_{n}=x$. Can we say that $\lim _{n \rightarrow \infty}a_{n}^q=(\lim _{n \rightarrow \infty}a_{n})^q=x^q$, where $q$ is a real number. Prove or disprove this result.

Here is what I have done:

proof: Since, $ \{ a_n \}_{n=m}^{\infty}$ is a convergent sequence. So, for every $\epsilon >0$, there exists a natural number $N \geq m$ so that, we have $$ |a_{n}-x|< \epsilon \; \forall n\geq N $$ Now, I have proven that $$ x^q-(x-\epsilon)^q\leq |a_{n}^q-x^q| \leq (x+\epsilon)^q-x^q $$ After this point I am not sure how to proceed. One way I can think of is using Newton's generalized binomial theorem to show that both upper bound and lower are function of $\epsilon$ (which can be easily derived) and we can make it as small as we want. Will this suffice the proof of the result?

  • $\begingroup$ If $f$ is a continuous function and $\lim_{n\to\infty}a_n=a$ then $$\lim_{n\to\infty}f(a_n)=f(a). $$ Here $f(x)=x^q$ which is continuous, so the result in question is true. $\endgroup$ – Math1000 Apr 29 '16 at 11:21
  • $\begingroup$ Yes, the result is true for this case but how to prove this as a general result. $\endgroup$ – MUH Apr 29 '16 at 12:24

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