Let $y$ be real, $n$ is natural, and $\varepsilon>0$. For some $\delta>0$, if $u$ is real and $|u-y|<\delta$, then $|u^{n}-y^{n}|<\varepsilon$ I've almost got this problem solved, and I need a(some) pointer(s) as to finishing this problem up (which will be greatly appreciated [!]).  Here is the full, problem statement in its exact form.
Problem: Given $y\in\mathbb{R}$, $n\in\mathbb{N}$, and $\varepsilon>0$, show that for some $\delta>0$, if $u\in\mathbb{R}$ and $|u-y|<\delta$ then $|u^{n}-y^{n}|<\varepsilon$.
Preliminary Notes/Work: Now, the question additionally provides a hint to prove the inequality for $n=1$ and $n=2$, then to use induction on $n$ and use the identity of $u^{n}-y^{n}=(u-y)(u^{n-1}+yu^{n-2}+\cdots+y^{n-2}u+y^{n-1})$. I've made it as far as defining the induction hypothesis for some $n$ and I've used the identity to work the inequalities in order to show the statement holds for $n$, but I can't find a way out from there.
So, proceeding by induction, we have for the case when $n=1$ that we can simply let $\delta=\varepsilon$. Then, letting $y\in\mathbb{R}$, and taking $\varepsilon>0$, let $\delta=\varepsilon>0$. When $u\in\mathbb{R}$ this implies that when $|u-y|<\delta=\varepsilon$, then clearly $|u^{1}-y^{1}|=|u-y|<\varepsilon$.
For the case when $n=2$, we let $\delta=\min\bigg\{1,\dfrac{\varepsilon}{2|y|+1}\bigg\}$. Now, when $u\in\mathbb{R}$ and $|u-y|<\delta$, then $|u-y|<1$ in particular. Further, by the Reverse Triangle Inequality, we then have that $|u|-|y|\leq |u-y|<1$. Adding $|2y|$ to both sides yields that $|u|+|y|\leq |u-y|+|2y|<1+|2y|$. Again by the Triangle Inequality we have that $|u+y|\leq |u|+|y|<2|y|+1$, and so $|u+y|<2|y|+1$. We now compute $|u^{2}-y^{2}|=|u+y||u-y|<(2|y|+1)\dfrac{\varepsilon}{2|y|+1}=\varepsilon$. Thus, the result hold also when $n=2$, and we can conclude the base case holds.
For the inductive hypothesis we have that for $y\in\mathbb{R}$, $n-1\in\mathbb{N}$, and $\varepsilon>0$, there is some $\delta>0$ such that if $u\in\mathbb{R}$ and $|u-y|<\delta$ then $|u^{n-1}-y^{n-1}|<\varepsilon$. We need to show this result holds for $n$.
I've been messing around with this, and I keep getting nowhere. We have that when $|u-y|<\delta$ then:
$|u^{n}-y^{n}|=|u-y||u^{n-1}+yu^{n-2}+\cdots+y^{n-2}u+y^{n-1}|$
$~~~~~~~~~~~~~~~~\!<\delta|u^{n-1}+yu^{n-2}+\cdots+y^{n-2}u+y^{n-1}|$.
I just need to find the $\delta$ in this case, and, at first, I wanted to let $\delta=\dfrac{\varepsilon}{u^{n-1}+yu^{n-2}+\cdots+y^{n-2}u+y^{n-1}}$, but (I think) the problem statement asserts that $\delta$ can't depend on $u$, or can it?  If it can, then I'm done...but if not , which $\delta$ will work?  Please note, this problem is a homework problem, and if one decides not to provide a solution (which I completely understand), then please provide suggestions, recommendations, hints, etc. I figured since I'm almost done with the problem, providing a solution wouldn't hurt, but anything will be GREATLY appreciated!  Thank you for your time reading this.
 A: 
The way you're trying to do it doesn't require induction

$\delta$ can't depend on $u$ but it can depend on $y$. Note that $|u|<|u-y|+|y|<|y|+\delta$.
$$
\Big{|}\sum_{j=0}^{n-1}y^ju^{n-1-j}\Big{|}<\sum_{j=0}^{n-1}|y|^j|u|^{n-1-j}<\sum_{j=0}^{n-1}|y|^{j}(|y|+\delta)^{n-1-j}={(|y|+\delta)^{n}-|y|^{n}\over\delta} 
$$
So we want to have $$\delta{(|y|+\delta)^{n}-|y|^{n}\over\delta}<\varepsilon
\implies (|y|+\delta)^{n}<{\varepsilon}+|y|^n\\
\implies\delta<\left({\varepsilon}+|y|^n\right)^{1\over n}-|y|=:K$$
So picking $\delta=0.5K$ suffices.

The inductive way

$$|u^{n}-y^n|=|(u^{n-1}-y^{n-1})(u+y)+uy^{n-1}-yu^{n-1}|\\<|u^{n-1}-y^{n-1}||u+y|+|uy||y^{n-2}-u^{n-2}|$$
Now there exists $\delta_1,\delta_2>0:|u^{n-1}-y^{n-1}|<{0.5\varepsilon\over2|y|+1}$ whenever $|u-y|<\delta_1$ and $|u^{n-2}-y^{n-2}|<{0.5\varepsilon\over|y|(|y|+1)}$ whenever $|u-y|<\delta_2$. If we pick $0<\delta<\min\{\delta_1,\delta_2,1\}$ those inequalities still hold if $|u-y|<\delta$, moreover we have $|u|<|y|+\delta<|y|+1$. So
$$
|u-y|<\delta\implies|u^{n}-y^n|<0.5\varepsilon{|u+y|\over2|y|+1}+0.5\varepsilon{|u||y|\over|y|(|y|+1)}\\<0.5\varepsilon{|u|+|y|\over2|y|+1}+0.5\varepsilon{|u|\over(|y|+1)}<0.5\varepsilon+0.5\varepsilon=\varepsilon
$$

Book's method(?)

$$
|u^{m-j}y^{j}+u^jy^{m-j}|=|u|^j|y|^j|u^{m-j}+y^{m-j}|<|u|^j|y|^j(|u^{m-j}-y^{m-j}|+2|y|^{m-j})
$$
For $j=0,1,..,n-1$ we can find $\delta_j>0:|u^{n-1-j}-y^{n-1-j}|<{1\over n\{(|y|+1)|y|\}^j}$ whenever $|u-y|<\delta_j$. Pick $0<\delta<\min(\{\delta_j:0\le j\le n-1\}\cup\{1\})$.
$$
|u-y|<\delta\implies|u^n-y^n|<\delta\Big{|}\sum_{j=0}^{n-1}u^{n-1-j}y^j\Big{|}<\delta\sum_{j=0}^{\lfloor {n\over 2}\rfloor}|u^{n-1-j}y^j+y^{n-1-j}u^j|\\
<\delta\sum_{j=0}^{\lfloor {n\over 2}\rfloor}(|u||y|)^j\left({1\over n\{(|y|+1)|y|\}^j}+2|y|^{n-1-j}\right)\\
<\delta\sum_{j=0}^{\lfloor {n\over 2}\rfloor}\left({1\over n}+2|u|^j|y|^{n-1}\right)
$$
Now $$\sum_{j=0}^{\lfloor {n\over 2}\rfloor}|u|^j|y|^{n-1}<|y|^{n-1}\sum_{j=0}^{\lfloor {n\over 2}\rfloor}(|y|+1)^j=:S$$
Therefore
$$
|u^n-y^n|<\delta\left({{\lfloor {n\over 2}\rfloor}+1\over n}+2S\right)<\delta(1+2S)
$$
So picking $0<\delta<\min(\{\delta_j:0\le j\le n-1\}\cup\{1,{\varepsilon\over1+2S}\})$ suffices.
A: Let $n\in\mathbb Z_{>0}$ be given and let $f$ be the real-valued function defined by $f(x):=x^n.$ Then $f$ is clearly continuous, which means that for all $x\in\mathbb R$ and $\forall\varepsilon>0$ there is some $\delta>0$ such that $|f(z)-f(x)|=|z^n-x^n|<\varepsilon$ whenever $z\in\mathbb R$ and $|z-x|<\delta.$
