Limit problem (by definition I think)

In summary: You need an extra factor of 2 somewhereOh wait, I see. They want you to use the derivative. You need to use the second one. I was assuming it was a typo and we're supposed to use the first.In summary, the value of the limit is 24x.
  • #1
terryds
392
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
 
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  • #2
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.
 
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  • #3
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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  • #4
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 !
 
  • #5
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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What is a limit problem?

A limit problem is a type of mathematical problem that involves finding the value that a function approaches as the input approaches a certain value. It is used to analyze the behavior of a function near a specific point.

What is the definition of a limit?

The limit of a function f(x) as x approaches a value c is the value that f(x) gets closer and closer to as x gets closer and closer to c. It is denoted by the symbol lim f(x) = L, where L is the limit value.

How do you solve a limit problem?

To solve a limit problem, you first need to determine the limit expression, which includes the function and the value that the input is approaching. Then, you can use different techniques such as direct substitution, factoring, or L'Hôpital's rule to evaluate the limit.

What is the importance of limit problems?

Limit problems are important in calculus and other areas of mathematics because they allow us to analyze the behavior of a function and make predictions about its values. They also help us understand the concept of infinity and how functions behave at specific points.

Can you give an example of a limit problem?

One example of a limit problem is finding the limit of the function f(x) = (x^2 - 3x + 2)/(x - 1) as x approaches 1. By plugging in 1 for x, we get an indeterminate form of 0/0. Using algebraic manipulation or L'Hôpital's rule, we can evaluate the limit to be 2.

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