# Prove that this limit equals $5f'(x)$ [closed]

$$\lim_{h\to0}\frac{f(x+2h)-f(x-3h)}{h} =5f'(x)$$

From the basic form $$f(x+h)-f(x)$$ I used $$x-3h$$ as $$x$$ and $$x+2h$$ as $$x+h$$ and used this to make the question like this.

• Welcome to Math SE. I've added our preferred formatting. Note in particular a derivative uses an apostrophe, not a backtick.
– J.G.
Commented Sep 8, 2022 at 20:39
• Hint:$$\frac{f(x+2h)-f(x-3h)}{h}=2\frac{f(x+2h)-f(x)}{2h}+3\frac{f(x-3h)-f(x)}{-3h}.$$
– J.G.
Commented Sep 8, 2022 at 20:42

The derivative is the slope of the tangent line, which is the limiting value of slopes of secant lines as two points converge on the graph of the function. With a bit of manipulation, we can make your limit look like such a limit.

If you consider the secant line connecting points on the graph above $$x-3h$$ and $$x+2h$$, the slope is $$\frac{f(x+2h) - f(x-3h)}{(x+2h) - (x-3h)} = \frac{f(x+2h) - f(x-3h)}{5h} = \frac{1}{5} \cdot \frac{f(x+2h) - f(x-3h)}{h}.$$ Now, notice that as $$h \to 0$$, each of $$x-3h \to x$$ and $$x+2h \to x$$, and the difference $$5h \to 0$$ as well, so \begin{align} \lim_{h \to 0} \frac{f(x+2h) - f(x-3h)}{h} &= 5 \cdot \lim_{h \to 0} \frac{f(x+2h) - f(x-3h)}{5h} \\[3pt] &= 5 \cdot \lim_{h \to 0} \frac{f(x+2h) - f(x-3h)}{(x+2h) - (x-3h)} \\ &= 5 \cdot f'(x) \end{align}

• Crystal clear. +1 Commented Sep 8, 2022 at 21:49

$$\lim_{h\to 0}\frac{f(x+2h)-f(x-3h)}{h}=\lim_{h\to 0}\frac{f(x+2h)-f(x)+f(x)-f(x-3h)}{h}=\lim_{h\to 0}\frac{f(x+2h)-f(x)}{h}-\lim_{h\to 0}\frac{f(x-3h)-f(x)}{h}$$

Consider $$\lim_{h\to 0}\frac{f(x+2h)-f(x)}{h}$$.

Let $$h=\frac{t}{2}$$, then $$t\to 0$$ as $$h\to 0$$: $$\lim_{h\to 0}\frac{f(x+2h)-f(x)}{h}=\lim_{t\to 0}\frac{f(x+t)-f(x)}{\frac{t}{2}}=2\lim_{t\to 0}\frac{f(x+t)-f(x)}{t}=2f'(x)$$

Similarly $$\lim_{h\to 0}\frac{f(x-3h)-f(x)}{h}$$.

Let $$h=-\frac{t}{3}$$, then $$t\to 0$$ as $$h\to 0$$: $$\lim_{h\to 0}\frac{f(x-3h)-f(x)}{h}=\lim_{t\to 0}\frac{f(x+t)-f(x)}{-\frac{t}{3}}=-3\lim_{t\to 0}\frac{f(x+t)-f(x)}{t}=-3f'(x)$$

Thus, $$\lim_{h\to 0}\frac{f(x+2h)-f(x)}{h}-\lim_{h\to 0}\frac{f(x-3h)-f(x)}{h}=2f'(x)-(-3f'(x))=5f'(x)$$.