Verified answer

I'll assume that A, B are positive constants; otherwise this is trivially true.

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Note that

|f(x) g(x) - f(c) g(c)|

= |f(x) g(x) - f(c) g(x) + f(c) g(x) - f(c) g(c)|

≤ |f(x) g(x) - f(c) g(x)| + |f(c) g(x) - f(c) g(c)|

= |g(x)| |f(x) - f(c)| + |f(c)| |g(x) - g(c)|

≤ B |f(x) - f(c)| + A |g(x) - g(c)|, since |f(x)| ≤ A and |g(x)| ≤ B for all x in R.

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Here's the proof.

Given ε > 0, since f and g are continuous at x = c, there exists δ₁, δ₂ > 0 such that

|x - c| < δ₁ ==> |f(x) - f(c)| < ε/(2B) and |x - c| < δ₂ ==> |g(x) - g(c)| < ε/(2A).

Let δ = min {δ₁, δ₂}. Then, |x - c| < δ ==> |f(x) - f(c)| < ε/(2B) and |g(x) - g(c)| < ε/(2A).

So, |f(x) g(x) - f(c) g(c)|

= |f(x) g(x) - f(c) g(x) + f(c) g(x) - f(c) g(c)|

≤ |f(x) g(x) - f(c) g(x)| + |f(c) g(x) - f(c) g(c)|

= |g(x)| |f(x) - f(c)| + |f(c)| |g(x) - g(c)|

≤ B |f(x) - f(c)| + A |g(x) - g(c)|

< B (ε/(2B)) + A (ε/(2A)

= ε, as required.

That is, f(x)g(x) is continuous at x = c.

I hope this helps!

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