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?
