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

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.

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

2 Answers 2

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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}

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  • $\begingroup$ Crystal clear. +1 $\endgroup$
    – Randall
    Commented Sep 8, 2022 at 21:49
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$\lim_{h\to 0}\frac{f(x+2h)-f(x-3h)}{h}=\lim_{h\to 0}\frac{f(x+2h)-f(x)+f(x)-f(x-3h)}{h}=\lim_{h\to 0}\frac{f(x+2h)-f(x)}{h}-\lim_{h\to 0}\frac{f(x-3h)-f(x)}{h}$

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

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