Prove by induction $|u - y| < \delta \Rightarrow |u^{n} - y^{n}| < \epsilon$ Specifically, show that $\forall u \in \mathbb{R}$, $\forall n \in \mathbb{N}$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall y \in \mathbb{R} \quad |u - y| < \delta \Rightarrow |u^{n} - y^{n}| < \epsilon$
I am trying to see if I am understanding this correctly. I have written out a proof in as much detail as I think is necessary.  I feel shaky when it comes to using the $P(n)$ case to help prove $P(n+1)$.
Proof.
Base case n=1 is trivial.  We will only proceed with the inductive step n+1.
For convenience suppose $|u| > |y|$.  We expand $P(n+1)$
\begin{align*}
|u^{n+1} - y^{n+1}| & = |(u-y)(u^{n} + u^{n-1}y + ... + y^{n})| \\
& \leq |u-y|(|u^{n}| + |u^{n-1}y| + ... + |y^{n}|) \\
& \leq |u-y| \cdot (n+1) \cdot \max(\{|u|,|y|\})^{n} \\
& < \delta \cdot (n+1) \cdot |u|^{n}
\end{align*}
For $P(n)$ the statement holds true so lets first expand on this case to be able to proceed with case $P(n+1)$.
\begin{align*}
|u^{n} - y^{n}| & = |(u-y)(u^{n-1} + u^{n-2}y + ... + y^{n-1})| \\
& \leq |u-y|(|u^{n-1}| + |u^{n-2}y| + ... + |y^{n-1}|) \\
& \leq |u-y| \cdot n \cdot \max(\{|u|,|y|\})^{n-1} \\
& < \delta \cdot n \cdot |u|^{n-1} \\
& < \delta \cdot (n+1) \cdot |u|^{n-1}
\end{align*}
Our choice of delta will depend on the value of $|u|$ and $\epsilon$.  If $|u| \leq 1 \Rightarrow |u|^{n} \leq 1$ for any $n \in \mathbb{N}$ (it does not matter if it is $n-1$ or $n$), such that
\begin{align*}
\delta \cdot (n+1) \cdot |u|^{n} \leq \delta \cdot (n+1) \cdot 1
\end{align*}
and so we pick $\delta = \frac{\epsilon}{n+1}$.  If $|u| > 1 \Rightarrow |u|^{n} > |u|^{n-1}$ such that
\begin{align*}
\delta \cdot (n+1) \cdot |u|^{n-1} < \delta \cdot (n+1) \cdot |u|^{n}
\end{align*}
and so we pick $\delta = \frac{\epsilon}{(n+1) \cdot |u|^{n}}$.
In any of the two cases we can choose a delta such that.
\begin{align*}
|u - y| < \delta \Rightarrow |u^{n} - y^{n}| < \delta \cdot (n+1) \cdot |u|^{n} = \epsilon
\end{align*}
a statement that holds true since $P(n)$ holds true.
Having found the necessary delta, we return to $P(n+1)$.  We choose $\delta = \frac{\epsilon}{n+1}$ if $|u| \leq 1$ or $\delta = \frac{\epsilon}{(n+1) \cdot |u|^{n}}$ if $|u| > 1$ such that the following is holds true
\begin{align*}
|u^{n+1} - y^{n+1}| & < \delta \cdot (n+1) \cdot |u|^{n} = \epsilon \\
|u^{n+1} - y^{n+1}| & < \epsilon
\end{align*}
Any feedback is appreciated!  Thanks!
 A: You don't seem to use the truth of $P(n) $ to establish the truth of $P(n+1)$. Rather you establish the truth of both $P(n) $ and $P(n+1)$ independently using same approach.
You can however use induction in following manner. It is obvious that $P(1)$ is true. Let's assume the truth of $P(n) $. Thus given any $\epsilon>0$ and $u\in\mathbb{R} $ there is a $\delta>0$ such that if $|u-y|<\delta$ then $|u^n-y^n|<\epsilon $.
Now consider the expression $$|u^{n+1}-y^{n+1}|\leq |u^{n+1}-u^ny|+|u^ny-y^{n+1}|\\\leq |u^n||u-y|+|y||u^n-y^n|$$ If $|u-y|<1$ then $|y|<|u|+1$ and hence by above inequality we have $$|u^{n+1}-y^{n+1}|\leq (|u^n|+1)|u-y|+(|u|+1)|u^n-y^n|$$ Let $\epsilon>0$ be given. Then by truth of $P(n) $ there is a $\delta_1>0$ such that if $|u-y|<\delta_1$ then $|u^n-y^n|<\epsilon/(2(|u|+1))$. Let $$\delta=\min\left(1,\delta_1,\frac{\epsilon}{2(|u^n|+1)}\right)$$ then for $|u-y|<\delta$ we have $$|u^{n+1}-y^{n+1}|<(|u^n|+1)\cdot\frac{\epsilon}{2(|u^n|+1)}+(|u|+1)\cdot\frac{\epsilon}{2(|u|+1)}=\epsilon$$ and this establishes the truth of $P(n+1)$.
