If \(f(x)\) is differentiable at \(b\), then \(f'(b)\) exists:
\[f'(b) = \lim_{x→b} \frac{f(x) - f(b)}{x - b}\]
We want to show that \(f(x)\) is continuous at \(b\) by showing that \(\lim_{x→b} f(x) = f (b)\). We can rewrite \(\lim_{x→b}\) as:
\[\lim_{x→b} f(x) = \lim_{x→b}(f(x)-f(b)+f(b))\]
Multiply and divide \(f(x) - f(b)\) by \(x - b\):
\[= \lim_{x→b} \left( \frac{f(x)-f(b)}{x-b}(x-b) + f(b) \right)\]
Using the sum law and the product law:
\[= \left(\left(\lim_{x→b} \frac{f(x)-f(b)}{x-b}\right) (\lim_{x→b} (x-b))\right) + \lim_{x→b}f(b)\]
If we evaluate it:
\[= (f'(b) (0) ) + \lim_{x→b}f(b) = \lim_{x→b}f(b)\]
Since \(\lim_{x→b}f(x)=f(b)\), we conclude that \(f\) is continuous at \(b\).