Differentiable proof

Question

Let $$f:(a,b)\to \mathbb R$$ be differentiable at $x_0\in (a,b)$. Prove that $$\lim_{x\to x_0}\frac{xf(x_0)-x_0f(x)}{x-x_0}=f(x_0)-x_0f'(x_0)\,.$$ Is this as simple as multiplying by a conjugate and reducing? I tried but I got lost in the simplifying.

Answer 1

Note that \begin{align*} \lim_{x \to x_0} \frac{xf(x_0)-x_0f(x)}{x-x_0} &= \lim_{x \to x_0} \frac{xf(x_0) - x_0f(x_0) + x_0f(x_0)-x_0f(x)}{x-x_0}\\ &= f(x_0)\lim_{x \to x_0} \frac{x - x_0}{x-x_0} + \lim_{x \to x_0} \frac{x_0f(x_0)-x_0f(x)}{x-x_0}\\ &= f(x_0) - x_0\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}\\ &= f(x_0) - x_0f^\prime(x_0). \end{align*} It's also helpful to remember that this type of "addition/substraction" trick is useful when dealing with such problems!