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I want a $\epsilon-\delta$ proof for the continuity of $f(x)=x^n$ around a point $a$ on the domain of this function. Here's my sketch:

$$|f(x) - f(a)| = |x^n - a^n| = |x-a||x^{n-1} + x^{x-2}a +\dotsb + xa^{n-2} + a^{n-1}|$$

For every pair $x^{n - j}a^{j-1} + x^{j-1}a^{n-j}$ we can write as

\begin{multline} x^{j-1}a^{j-1}\bigl((x^{n-2j +1} - a^{n-2j+1}) + 2a^{n-2j+1} \bigl)\\ = x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl) \end{multline}

for every $j = 1,2,..., n$

Since $|x-a|<\delta$:

$$\bigl|x^{j-1}a^{j-1}\bigl((x - a)(x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + 2a^{n-2j+1} \bigl)| < |x^{j-1}a^{j-1}| \bigl( \delta (x^{n-2j} + x^{n-2j -1}a +\dotsb + xa^{n-2j-1} + a^{n-2j}) + |2a^{n-2j+1}| \bigl)$$

From now on I can't think of anything that would help me to conclude, I can't find any pattern that would help me summarize it, I would be glad if you help me.

Thanks.

DMcMor
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João Hiroyuki
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    I would keep it simple. Prove with $\varepsilon, , \delta$ that the second factor tends to $n a^{n-1}$ and your first factor tends to $0$. Then use algebra limits. If you don't want to use algebra limit you can prove that there exist a $\delta$ such that $|x-a|< \frac{\varepsilon}{2na^{n-1}}$ – Manuel Jul 20 '17 at 14:37
  • You mean:

    $|x-a||x^{n-1} + x^{x-2}a +\dotsb + xa^{n-2} + a^{n-1}| < \delta |x^{n-1} + x^{x-2}a +\dotsb + xa^{n-2} + a^{n-1}|$

    Doing $x->a$, we have

    $\delta |na^{n-1}| < \epsilon$. Is that it? It's my first time studying analysis, I don't get the techniques yet...

     – João Hiroyuki Jul 20 '17 at 14:53
  • You don't want a proof by induction? 1 - Prove that $f(x) = x$ is continuous at $a$. 2 - Since x is continuous at a, so is $x\cdot x$, because it is the product of two continuous functions. 3 - And so on... – Quantaliinuxite Jul 20 '17 at 15:07
  • The book asks a $\epsilon-\delta$ proof, actually. It's the Introcution to Calculus and Analysis by Richard Courant. – João Hiroyuki Jul 20 '17 at 15:10

2 Answers2

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A few standard idioms (proofs left to you) will help.

  1. You may as well assume $\delta \leq 1$. (If $\delta > 0$ is arbitrary, you can replace it with $\min(\delta, 1)$.)

  2. If $x$ and $a$ are real numbers, then $$ |x - a| < \delta\quad\text{if and only if}\quad a - \delta < x < a + \delta. $$

  3. Consequently, if $|x - a| \leq 1$, then $|x| \leq 1 + |a|$.

    You don't need it here, but there's also a useful lower bound: If $|x - a| \leq 1$, then $|x| \geq 1 - |a|$ and $|x| \geq |a| - 1$. Succinctly, $|x| \geq \bigl||a| - 1\bigr|$.

  4. If $j$ and $m$ are integers such that $0 \leq j \leq m$, and if $|x - a| \leq 1$, then $$ |x^{j} a^{m-j}| = |x|^{j}\, |a|^{m-j} \leq (1 + |a|)^{m}. $$

  5. If $|x - a| \leq 1$, then the triangle inequality gives $$ |x^{n-1} + x^{x-2}a + \dots + xa^{n-2} + a^{n-1}| \leq n (1 + |a|)^{n-1}. $$

Andrew D. Hwang
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Hint:

Establish as a lemma that the product of two continuous function is continuous, using

$$|f(x)g(x)-f(a)g(a)|=|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|\\<|f(x)|\,|g(x)-g(a)|+|f(x)-f(a)|\,|g(a)|.$$