# Verification of an epsilon-delta proof

I'm following Michael Spivak's "Calculus". Here's the question:

$$0 < |x - 2| < \sin^2\left(\frac{\epsilon^2}{9}\right) + \epsilon\implies |f(x) - 2| < \epsilon$$

$$0 < |x-2| < \epsilon^2\implies |g(x) - 4| < \epsilon$$

Now, we are to prove that $$\lim_{x\to 2}(f\cdot g)(x) = 8$$. The approach I take is the following:

$$|f(x)g(x) - 8| = |f(x)g(x) - 2g(x) + 2g(x) - 8|\leq |g(x)||f(x) - 2| + 2|g(x) - 4|$$

Now, I set $$|g(x) - 4| < \frac{\epsilon}{4}$$. Then, the right summand becomes less than $$\frac{\epsilon}{2}$$. Now, as a consequence:

$$|g(x)| < \frac{\epsilon + 16}{4}$$ Thus, it is required that $$|f(x) - 2| < \frac{2\epsilon}{\epsilon+16}$$

However, the solution manual requires that

$$|f(x) - 2| < \min\left(1,\frac{\epsilon}{10}\right)\qquad |g(x) - 4| < \frac{\epsilon}{5}$$

So, my question is, are both answers valid? Or have I made a mistake?

• Where, in Spivak's Calculus, can that be found? Commented Jun 26 at 10:51
• @JoséCarlosSantos This is Problem 6 part (ii) of Chapter 5 (I'm referring to the 3rd edition) Commented Jun 26 at 10:52
• Well, then you made a mistake. It should be $|g(x)-4|<\varepsilon$, rather than $|g(x)-2|<\varepsilon$. Commented Jun 26 at 10:57
• @JoséCarlosSantos Ah, I apologize. However, my entire proof uses $|g(x) - 4| < \epsilon$ not $|g(x) - 2| < \epsilon$. That was just a typing error Commented Jun 26 at 10:59
• I see nothing wrong with your approach. Commented Jun 26 at 11:08