Limit problem (by definition I think)

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Homework Help Overview

The discussion revolves around evaluating the limit of a function defined as ##f(x)=12x^2-5##, specifically the expression ##\lim_{h\rightarrow 0}\frac{f(x+2h)-f(x-3h)}{6h}##. Participants explore the relationship of this limit to the definition of a derivative.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt to relate the limit to the derivative, expressing it in terms of ##f'(x)##. Others suggest directly evaluating the limit instead of complicating the approach. There is also discussion about potential typos in the problem statement that could affect the outcome.

Discussion Status

The conversation is ongoing, with some participants providing alternative evaluations of the limit and questioning the original problem setup. A few have arrived at different answers, indicating a lack of consensus on the correct approach or interpretation.

Contextual Notes

Participants note that the answer derived from their evaluations does not match the provided options, raising concerns about possible errors in the problem statement.

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