Prove the scalar multiple property of limits: limx->c [bf(x)]=bL

It needs to be the actual proof not an example, which is why I am stuck =(

Given ε > 0, we need to find δ > 0 such that 0 < |x - c| < δ ==> |b f(x) - bL| < ε.
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If b = 0, this is trivial to prove.
Given ε > 0, any δ > 0 will work: 0 < |x - c| < δ ==> |0 * f(x) - 0 * L| = 0 < ε.

Now suppose that b ≠ 0, and let ε > 0 be given

Since lim(x→c) f(x) = L, there exists δ > 0 such that 0 < |x - c| < δ ==> |f(x) - L| < ε/|b|.
Then, 0 < |x - c| < δ ==> |bf(x) - bL| = |b| |f(x) - L| < |b| * ε/|b| = ε.

Hence, lim(x→c) b f(x) = bL.
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I hope this helps!