Why do we need to add an extra upper bound in this case?

Question

In the proof of Baby Rudin Theorem 1.21 (I don't quote it here since it's accessible online and also since it might clutter an already long question), we use the following idea:

"If you change a number by just a little, then its $n$th power will be changed by just a little; in particular, assuming $y^n < x$, then if you change $y$ by a small enough value $h$, then $(y+h)^n$ will still be $< x$". To make precise the statement that "if you change a number by just a little, then its $n$th power will be changed by just a little", Rudin notes the formula which describes the difference in the $n$th power of two numbers "as a function of" the difference between the two numbers themselves: $$b^n - a^n = (b - a)(b^{n-1} + b^{n-2}a + ... + a^{n-1}).$$ The factor $b - a$ on the left of the RHS shows that if $b$ is "just a little" bigger than $a$, then the difference $b^n - a^n$ will also be "small". But how small it is is affected by the other factor, $b^{n-1} + b^{n-2}a + ... + a^{n-1}$. To make $b^n - a^n$ less than some value, we have to make $b - a$ less than that value divided by some upper bound for $b^{n-1} + b^{n-2}a + ... + a^{n-1}$; an upper bound that will do as shown in the Lemma (this is in the Rudin proof, I call it a lemma) is $nb^{n-1}$ if $b > a > 0$. We now want to apply this to make $(y+h)^n - y^n$ less than the distance from $y^n$ to $x$. To do this we choose the difference between $y+h$ and $y$ to be smaller than $x - y^n$ divided by an upper bound for $nb^{n-1}$. But note that with $b = y+h$ we would be saying choose $h$ such that $(y+h)-y = h < \frac{x - y^n}{n(y+h)^{n-1}}$.

My question is, why can't we choose $h$ as at the end here? I think this is a more general question about how we can "choose" numbers. In the actual proof, we use the further requirement on $h$ that $h<1$, so that $n(y+h)^{n-1}< n(y+1)^{n-1}$ and we then choose $h < \frac{x - y^n}{n(y+1)^{n-1}}$ to get the desired inequality. But why was this last step necessary? It was certainly convenient, but would it have been wrong to skip it and just demand an $h$ as before my bolded question?

Answer 1

The inequality $$h<\frac{x-y^n}{n(y+h)^{n-1}}\tag1$$ contains $h$ on both sides. So it's not obvious that an $h$ can be found to satisfy the inequality. OTOH the quantity $\frac{x-y^n}{n(y+1)^{n-1}}$ is positive, so it is clear that one can find an $h>0$ to satisfy $$h<\frac{x-y^n}{n(y+1)^{n-1}}.\tag2$$ So you could demand $h$ to satisfy (1) but you'd have to expend effort explaining why such an $h$ exists. Better to save the reader the effort and jump to (2), which still manages to achieve the result $(y+h)^n-y^n<x-y^n$.