# Limit problem (by definition I think)

1. Apr 12, 2016

### terryds

1. The problem statement, all variables and given/known data
$f(x)=12x^2-5$
The value of $\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x-3h)}{6h}$ is ...

A. 8x
B. 10x
C. 12x
D.18x
E. 24x

2. Relevant equations

$f'(x) = \lim_{h->0}\frac{f(x+h)-f(x))}{h}$

3. The attempt at a solution

Looking at the problem question, it seems that it's kinda like the definition of derivative
But, it's different.
This is my attempt.

$\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x)}{2h}*\frac{2h}{6h}-\frac{f(x-3h)-f(x)}{3h}*\frac{3h}{6h}$
$= f'(x) * 1/3 - f'(x) * 1/2$
$= f'(x)*(-1/6)$
$= 24x (-1/6)$
$= -4x$

2. Apr 12, 2016

### PeroK

The derivative is not relevant. Why not just evaluate the limit?

I didn't check where you went wrong, but here is no need to complicate things. Just evaluate the limit as you see it.

PS The answer I get isn't on the list.

Last edited: Apr 12, 2016
3. Apr 12, 2016

### Ssnow

In order to have the derivative you must have $+\frac{f(x-3h)-f(x)}{-3h}$ so you have $24x(1/3+1/2)$ ...

4. Apr 12, 2016

### terryds

Hmm.. By just evaluating the limit, I get the answer 20x as the answer

Yeah, I missed that one
$24x (\frac{1}{3}+\frac{1}{2}) = 24 * \frac{5}{6} = 20x$

Thanks for your help @PeroK and @Ssnow !

5. Apr 12, 2016

### SammyS

Staff Emeritus
Are you sure there's not a typo? It could be either of the following.
$\displaystyle \frac{f(x+2h)-f(x-3h)}{5h}$​
.
$\displaystyle \frac{f(x+3h)-f(x-3h)}{6h}$​