Limit problem (by definition I think)

  • Thread starter terryds
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  • #1
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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
 

Answers and Replies

  • #2
PeroK
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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

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.
 
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  • #3
Ssnow
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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)## ...
 
  • #4
392
13
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

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 !
 
  • #5
SammyS
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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
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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