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 $$\frac{3}{5}\lim_{k\to 0}\frac{f(t+k)-f(t)}{k}=\frac{3}{5}f'(t)=\frac{3}{5}f'(x_0-h)$$

  • $\begingroup$ 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

$$ \dfrac{f(x_0 + 2h) - f(x_0 - h)}{5h} = \dfrac{2}{5}\dfrac{f(x_0 + 2h) - f(x_0)}{2h} + \frac{1}{5}\dfrac{f(x_0 - h) - f(x_0)}{-h} $$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.