How do I choose the appropriate delta for this epsilon - delta proof

Question

I'm trying to prove that limit $\lim_{x \rightarrow 0} (4 + cx\sin(\frac{1}{x})) = 4$ for some arbitrarily large $c$ using the epsilon-delta definition of a limit. The nature of the function has me stuck. Thanks

Answer 1

The goal is to show $\forall \epsilon > 0, \exists \delta > 0$ s.t. $0 < |x - 0| = |x| < \delta \implies \left| 4 + cx \sin \left( \frac{1}{x} \right) - 4 \right| = \left| cx \sin \left( \frac{1}{x} \right) \right| < \epsilon$.¹

The exposition below is informal in that one would typically begin the proof by proposing some expression for $\delta$ and showing that it produces the desired conclusion. This sketch shows how one might come up with such $\delta$; the proof can easily be constructed by "reversing" this sketch.

Observe that

$$ \begin{align*} \left| cx \sin \left( \frac{1}{x} \right) \right| &= |c||x|\left| \sin \left( \frac{1}{x} \right) \right| \\ &\leq |c| |x| \\ &<|c| \delta, \end{align*} $$

where the first inequality follows from the fact that $\sin$ is a bounded function with maximum 1.

We want $|c|\delta \leq \epsilon$. Therefore, we can choose $\delta = \epsilon / |c|$. This yields the desired claim.

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¹ The $0< |x|$ condition is necessary since by definition, limits only consider the "punctured neighborhood," which in this case is $N_\delta(0)$; removing the LHS of the inequality alters the definition to that of continuity. In particular, $\sin \left(\frac{1}{x}\right)$ is not continuous at $x = 0$, as the function is undefined at the point.