Limits, finding $\alpha$ so my function is continuous Let $g(x)=\dfrac{f(x)-f(c)}{x-c}+\alpha(x-c)$ if $x\neq c$ and $g(x)=f'(c)$ if $x=c$.
Find $\alpha$ such that $g'(x)$ is continuous.
I started with the definition of $g'(x)=\lim\limits_{x\to c}\dfrac{g(x)-g(c)}{x-c}$, with some steps later I don't know how to proceed.
 A: We're given
$$g(x)=\begin{cases}\cfrac{f(x)-f(c)}{x-c}+\alpha(x-c)\;,\,&x\neq c\\{}\\
f'(c)\;,\,&x=c\end{cases}$$
Begin by finding $\;g'(x)\;$:
$$\begin{align*}&x\neq c:\;\; g'(x)\stackrel{\text{diff. of fraction}}=\frac{(x-c)f'(x)-(f(x)-f(c))}{(x-c)^2}+\alpha\\{}\\
&x=c:\;\;g'(c)=f''(c)\end{align*}$$
Assuming $\;f',f''\;$ is continuous at  $\;x=c\;$  (it is not given but I don't think we can do without this), we have no problem with continuity $\;g'\;$ except perhaps at $\;x=c\;$ , so it must be that
$$f''(c)=g''(c)=\lim_{x\to c}g'(x)=\lim_{x\to c}\frac{(x-c)f'(x)-(f(x)-f(c))}{(x-c)^2}+\alpha\stackrel{\text{l'H 1st. summand}}=$$
$$=\lim_{x\to c}\frac{f'(x)+(x-c)f''(x)-f'(x)}{2(x-c)}+\alpha=\frac{f''(c)}2+\alpha$$
Thus, it must be that
$$f''(c)=\frac{f''(c)}2+\alpha\implies\alpha=\frac12f''(c)$$
A slightly different way: we in fact have that
$$x\neq c\implies g'(x)=\frac{f'(x)-f'(c)}{x-c}+\frac{f'(c)}{x-c}-\frac{f(x)-f(c)}{(x-c)^2}+\alpha$$
$$g'(c)=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}=\lim_{x\to c}\frac{f(x)-f(c)+\alpha(x-c)^2-(x-c)f'(c)}{(x-c)^2}\stackrel{l'H}=$$
$$=\lim_{x\to c}\frac{f'(x)+2\alpha(x-c)-f'(c)}{2(x-c)}\stackrel{l'H}=\lim_{x\to c}\frac{f''(x)+2\alpha}{2}=\frac12f''(c)+\alpha$$
