Limit problem (by definition I think)

  • Thread starter Thread starter terryds
  • Start date Start date
  • Tags Tags
    Definition Limit
Click For Summary
The limit problem involves evaluating the expression ##\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x-3h)}{6h}## for the function ##f(x)=12x^2-5##. Initial attempts to solve it using the derivative led to an incorrect answer not listed among the options. By directly evaluating the limit, the correct answer was found to be ##20x##, which is also not among the provided choices. There was a suggestion that a typo might exist in the problem, as alternative formulations could yield different results. The discussion emphasizes the importance of evaluating the limit directly rather than overcomplicating the process with derivatives.
terryds
Messages
392
Reaction score
13

Homework Statement


##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

Homework Equations



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

The Attempt at a Solution


[/B]
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##

But, it's not in the options. Please help
 
Physics news on Phys.org
terryds said:

Homework Statement


##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

Homework Equations



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

The Attempt at a Solution


[/B]
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##

But, it's not in the options. Please help

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:
  • Like
Likes blue_leaf77
In order to have the derivative you must have ##+\frac{f(x-3h)-f(x)}{-3h}## so you have ##24x(1/3+1/2)## ...
 
  • Like
Likes terryds
PeroK said:
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.

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

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

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

Thanks for your help @PeroK and @Ssnow !
 
terryds said:

Homework Statement


##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
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}##​

Both give answer E.
 
  • Like
Likes terryds

Similar threads

  • · Replies 40 ·
2
Replies
40
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 25 ·
Replies
25
Views
1K
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K