Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
You should be careful, since your $t$ also depends on $h$ which affects $k$, so it's not a constant. –  Calvin Lin Jan 18 '13 at 22:03

1 Answer 1

up vote 5 down vote accepted

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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