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Let $f$ be a differentiable function in $x_0$. Calculate the following $\lim$: $$\lim_{h\to 0}\frac{f(x_0+2h)-f(x_0-h)}{5h}$$

since we know from theory that $f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$, then

I said that $x_0-h=t$ and $x_0+2h=t+3h$ where $3h=k$ so $$\frac{3}{5}\lim_{k\to 0}\frac{f(t+k)-f(t)}{k}=\frac{3}{5}f'(t)=\frac{3}{5}f'(x_0-h)$$

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  • $\begingroup$ You should be careful, since your $t$ also depends on $h$ which affects $k$, so it's not a constant. $\endgroup$ – Calvin Lin Jan 18 '13 at 22:03
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$$ \dfrac{f(x_0 + 2h) - f(x_0 - h)}{5h} = \dfrac{2}{5}\dfrac{f(x_0 + 2h) - f(x_0)}{2h} + \frac{1}{5}\dfrac{f(x_0 - h) - f(x_0)}{-h} $$

Now take the limit as $h \to 0$.

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