Hammack: Example 13.4 -- proof that there is no limit for sin(1/x) as x goes to 0

Question

In section "13.3 Limits That Do Not Exist", of his "Book of Proof" (3rd Edition) Hammack gives:

Example 13.4 Prove that $ \displaystyle \lim_{x \to 0} \sin\left({\frac{1}{x}}\right)$ does not exist.

As $x$ approaches $0$, the number $\frac{1}{x}$ grows bigger, approaching infinity, so $\sin(\frac{1}{x})$ just bounces up and down, faster and faster the closer $x$ gets to $0$.

His proof appears to be a direct proof that it is not true that there is a limit -- but it seems to start as if it is a proof by contradiction. This bothers me because a "proof of the contrary" does not seem to be the same as "proof by contradiction".

Here is the proof as given:

Proof. Suppose for the sake of contradiction that $ \displaystyle \lim_{x \to 0} \sin\left({\frac{1}{x}}\right) = L$ for $L \in \mathbb{R}$. Definition 13.2 guarantees a number $\delta$ for which $0 < |x - 0| < \delta)$ implies $|\sin\left({\frac{1}{x}}\right) - L| < \frac{1}{4}$. Select $k \in \mathbb{N}$ large enough so that $\frac{1}{k\pi} < \delta$. As $0 < \left|\frac{1}{k\pi} - 0\right| < \delta$, we have $\left|\sin\left(\frac{1}{1/k\pi}\right)-L\right| < \frac{1}{4}$, and this yields $\left|\sin\left(k\pi\right)-L\right| = \left|0-L\right| = \left|L\right|< \frac{1}{4}$.

Next, take $l \in \mathbb{N}$ large enough so that $\frac{1}{\frac{\pi}{2} + 2l\pi} < \delta$, so we have $0 < \left|\frac{1}{k\pi} - 0\right| < \delta$, we have $\left|\sin\left(\frac{1}{\frac{1}{\frac{\pi}{2} + 2l\pi}}\right)-L\right| < \frac{1}{4}$, which simplifies to $\left|\sin\left(\frac{\pi}{2} + 2l\pi\right)-L\right| = \left|1-L\right| < \frac{1}{4}$.

Above we showed $\left|L\right|< \frac{1}{4}$ and $\left|1-L\right| < \frac{1}{4}$. Now apply the inequality (13.2) to get the contradiction $1 < \frac{1}{2}$, as $1 = |L+(1-L)| \le |L| + |1-L| < \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

Hammack also gives:

... in symbolic form Definition 13.2 says $\displaystyle \lim_{x \to c} f(x) = L$ if and only if

    $\forall \epsilon > 0, \exists \delta > 0, (0 < |x - c| < \delta) \implies (|f(x) - L| < \epsilon)$          (13.4)

I believe the contrary of that is:

    $\exists \epsilon > 0, \forall \delta > 0, (0 < |x - c| < \delta) \land \lnot(|f(x) - L| < \epsilon)$

So the above proof appears to prove the contrary. It chooses an $\epsilon$ and shows that for all $\delta$ it is not true that $|f(x) - L| < \epsilon$, no matter what value of $L \in \mathbb{R}$ might be. Proving the last part is by contradiction: it is proved that assuming $|f(x) - L| < \epsilon$ is true (for all $\delta$ and all $L$) leads to a contradiction.

So my actual questions are:

1. perhaps I have misread Hammack, and all his proofs of the non-existence of limits are indeed (implicitly) direct proofs of not-existence (proof of the contrary), but the last step of that is generally a proof by contradiction ?

Or:

2. perhaps I am mistaken in thinking that "proof of the contrary" and "proof by contradiction" are different ?