We are given that function $\mathbb{f}$ is differentiable at $\mathbb x$. We have to show that it is also continuous at $\mathbb x$.
Now, according to definition of differentiability,
$\forall \epsilon > 0, \exists \delta > 0$ such that
$$|t-x| < \delta \implies \left|\frac{f(t)-f(x)}{t-x} - L\right| < \epsilon$$
Now, $|x - y| \ge |x| - |y|$.
$$\therefore \left|\frac{f(t)-f(x)}{t-x}\right| - |L|< \epsilon$$
$$\therefore |f(t) - f(x)| < |t-x| * (\epsilon + |L|) < \delta*(\epsilon+|L|)$$
Now take $\delta' = \frac{\epsilon}{\epsilon + |L|}$
So, $\forall \epsilon > 0, \exists \delta' > 0$ such that, $$|f(t) - f(x)| < \delta'*(\epsilon + |L|)$$ $$\therefore |f(t) - f(x)| < \left(\frac{\epsilon}{\epsilon + |L|}\right) * (\epsilon + |L|)$$ $$\therefore |f(t) - f(x)| < \epsilon$$
So, we have proved that $$\lim_{t\to x} f(t) = f(x)$$
Hence, function is continuous at x. Is this proof valid ? Thanks.