Suppose $f$ is differentiable at $a$. Evaluate $\lim\limits_{h\to0} \frac{f(a+16h) - f(a+15h)}h$ Suppose $f$ is differentiable at $a$. Evaluate if possible
$$\lim\limits_{h\to0} \frac{f(a+16h) - f(a+15h)}h$$
$$\lim\limits_{h\to0} \frac{f(a+15h)}h - \lim\limits_{h\to0} \frac{f(a+15h)}h$$
which to me is basically just $f'(a)- f'(a) = 0$
or
I found
$$\lim\limits_{h\to0} \frac{f(a) + f'(a)16h + r(16h) - f(a) - f'(a)15h - r(15h)}h$$
which equals to $\lim\limits_{h\to0} \frac{16hf'(a) - 15hf'(a)}h$ since $\lim\limits_{h\to0} r(16h)/h = 0$
which gives me $16f'(a) - 15f'(a) = f'(a)$
I end up with either $f'(a)$ or zero but I don't know which one is correct.
 A: Rewrite (think about the definition of the derivative):
$$\begin{array}{rcl}
\displaystyle \frac{f(a+16h) - f(a+15h)}{h} & = & \displaystyle \frac{f(a+16h)  \color{red}{-f(a)} -  f(a+15h)\color{red}{+f(a)} }{h} \\ \\
 & = & \displaystyle \color{blue}{16}\frac{f(a+16h) - f(a)}{\color{blue}{16}h} - \color{green}{15}\frac{ f(a+15h)-f(a)}{\color{green}{15}h}
\end{array}$$
So taking the limits of $h \to 0$:
$$\begin{array}{rcl}
\displaystyle \lim_{h \to 0}  \frac{f(a+16h) - f(a+15h)}{h} & = & \displaystyle \color{blue}{16}\lim_{h \to 0}\frac{f(a+16h) - f(a)}{\color{blue}{16}h} - \color{green}{15}\lim_{h \to 0}\frac{ f(a+15h)-f(a)}{\color{green}{15}h} \\ \\
& = & \displaystyle \color{blue}{16}f'(a)-\color{green}{15}f'(a) \\ \\
& = & \displaystyle f'(a) 
\end{array}$$
If you want, you could change variables $t=16h$ (in first limit; with $t \to 0$ as $h \to 0$) and $u=15h$ (in the second limit; with $u \to 0$ as $h \to 0$)) to get the literal definition of the derivative.
A: In fact, using definition of the derivative, you have
$$\lim\limits_{h\to0} \frac{f(a+16h)-f(a)}h = 16 \lim\limits_{h\to0} \frac{f(a+16h)-f(a)}{16h} = 16 \lim\limits_{\delta\to0} \frac{f(a+\delta)-f(a)}\delta = 16 f'(a).$$
So using the first approach you suggested you will get $16f'(a)-15f'(a)=f'(a)$.
