Limit of $f(x)=\frac{x}{1+\sin^2x}$ as $x\to0$ and proof

Question

I computed the limit using the limit theorems and the answer is obviously $0$. So now I am attempting to prove it using the $\epsilon,\delta$ definition.

$f(x)=\frac{x}{1+\sin^2x}$

$|\frac{x}{1+\sin^2x}|<\epsilon$ if $|x|<\delta$

$|x|<\epsilon|1+\sin^2x|$

$|\sin x|<1$ $\implies$ $|\sin^2x|<1$ $\implies$ $1+|\sin^2x|<2$ $\implies$ $|1+\sin^2x|<1+|\sin^2x|<2$ $\implies$ $|1+\sin^2x|<2$ $\implies$ $\epsilon|1+\sin^2x|<2\epsilon$

Finally,

$|x|<\epsilon|1+\sin^2x|<2\epsilon$ $\implies$ $|x|<2\epsilon=\delta$

But when I plug $\epsilon=0.2$ then the $\delta=0.4$. If I put my $x=0.3<0.4$ then the answer is $0.3>0.2=\epsilon$.

What am I doing wrong here ?

Also sometimes Spivak assumes (for example in $|x^2-a^2|<\epsilon$ if $|x-a|<\delta$) that $|x-a|<1$ and chooses $\min(1,\frac{\epsilon}{2|a|+1})$ I cant understand how is it possible to make the assumption that $|x-a|<1=\delta$ if our objective IS to find $\delta$ to prove that this $\delta$ exists for every $\epsilon$. Isnt this circular reasoning ? Otherwise we can just choose that $\delta=\frac{\epsilon}{10000000}$ (10000000 is an arbitrary number that makes delta very small).

Also in the example $|x^2-a^2|<\epsilon$ if $|x-a|<\delta$ assuming by some magical means that $|x-a|<1$ then $|x|-|a|<|x-a|<1$ $\implies$ $|x|<1 + |a|$ $\implies$ $|x+a|<|x|+|a|<2|a|+1$ $\implies$ $\frac{1}{|x+a|}>\frac{1}{2|a|+1}$ $\implies$ $\frac{\epsilon}{|x+a|}>\frac{\epsilon}{2|a|+1}$. And since $|x-a|<\frac{\epsilon}{|x+a|} >\frac{\epsilon}{2|a|+1}$ isn't it WRONG to write $|x-a|< \frac{\epsilon}{2|a|+1}$ since we DONT know if $|x-a|< \frac{\epsilon}{2|a|+1}$ is true or not ?

Apologies for the long question but limits have been utterly confusing me for a long while now.

Answer 1

You wish to prove that $$\forall \varepsilon >0\,\exists \delta >0\,\forall x\in \mathbb R\left(|x|<\delta\implies |f(x)|<0\right).$$

Let $\varepsilon >0$ and set $\delta:=\varepsilon.$ Now take $x\in \mathbb R$ and assume $|x|<\delta\color{grey}{=\varepsilon}$.

One has $|f(x)|=\left|\dfrac {x}{1+(\sin(x))^2}\right|\leq |x|<\delta=\varepsilon$, as required.

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Regarding the Spivak example, from what I gather he is considering two cases, either $\delta =1$ is good enough, or it isn't (these cases depend on $\varepsilon$). If it is the second case, $\delta=\dfrac{\epsilon}{2|a|+1}$ is good enough. Either way $\delta =\min\left(1,\frac{\epsilon}{2|a|+1}\right)$ takes care of every thing.

Answer 2

In response to:

Finally,

$|x|<\epsilon|1+sin^2x|<2\epsilon$ $\implies$ $|x|<2\epsilon=\delta$

But when I plug $\epsilon=0.2$ then the $\delta=0.4$. If I put my $x=0.3<0.4$ then the answer is $0.3>0.2=\epsilon$.

What am I doing wrong here ?

What you're doing wrong is trying to reverse the direction of the implication from $\Longrightarrow$ to $\Longleftarrow$.

It's certainly true that if $|x|$ is less than $\epsilon|1+\sin^2x|$, then $|x|$ is also less than $2$ times $\epsilon$. For that matter, $|x|$ is less than a gazillion times $\epsilon$. But the converse need not be true.

What you should really do is ask not how big $\epsilon|1+\sin^2x|$ can be, but how small it can ever be. The answer to that is clearly $\epsilon$, since $\sin^2x\ge0$ for all $x$. Using that for $\delta$ -- i.e., letting $\delta=\epsilon$ -- we have

$$|x|\lt\delta\implies|x|\lt\epsilon\implies|x|\lt\epsilon|1+\sin^2x|\implies\left|{x\over1+\sin^2x}\right|\lt\epsilon$$