Limit Question using first principles

[Jamesaa1234]

Homework Statement

If f(a) = 0 and f'(a) = 6 find lim h -> 0 (f(a+h)/2h).

Homework Equations

lim h ->0 (f(a+h)-f(a))/h

The Attempt at a Solution

I found the ratio between the two equations.
(f(a+h)-f(a))/h / (f(a+h)/2h)
I found this to be 2. Is this step possible or can you not take the ratio for first principles?
Then, I did:
lim h-> 0 (2(f(a+h)/2h) = 6
lim h-> 0 (f(a+h)/2h) = 3 (is this step also possible or is this incorrect?)

 

[Ray Vickson]

Homework Statement

If f(x) = 0 and f'(x) = 6 find lim h -> 0 (f(a+h)/2h).

Homework Equations

lim h ->0 (f(x+h)-f(x))/h

The Attempt at a Solution

I found the ratio between the two equations.
(f(x+h)-f(x))/h / (f(a+h)/2h)
I found this to be 2. Is this step possible or can you not take the ratio for first principles?
Then, I did:
lim h-> 0 (2(f(a+h)/2h) = 6
lim h-> 0 (f(a+h)/2h) = 3 (is this step also possible or is this incorrect?)

The question makes no sense. Either you are given a specific value of x at which f(x) = 0 and f'(x) = 6, or else (if it is supposed to hold for all x) it is imposs ible.

 

[Jamesaa1234]

The question makes no sense. Either you are given a specific value of x at which f(x) = 0 and f'(x) = 6, or else (if it is supposed to hold for all x) it is imposs ible.

I added a instead of x. I edited and fixed it up, does it make sense now?

 

[Ray Vickson]

I added a instead of x. I edited and fixed it up, does it make sense now?

No. If you mean that f'(x) = 6 for all x, then f(x) depends on x; it cannot be the constant f(x) = 0 for all x. You CAN have f(x₀) = 0 at a single point x₀, and have f'(x) = 6 for all x, but that is not what you wrote.

 

[Jamesaa1234]

No. If you mean that f'(x) = 6 for all x, then f(x) depends on x; it cannot be the constant f(x) = 0 for all x. You CAN have f(x₀) = 0 at a single point x₀, and have f'(x) = 6 for all x, but that is not what you wrote.

Okay, but am i allowed to take the ratio two lim equations if i do not write the lim sign?
Also its f prime of a = 6 just to clarify.

 

[Mark44]

Okay, but am i allowed to take the ratio two lim equations if i do not write the lim sign?
Also its f prime of x = 6 just to clarify.

You can divide one limit expression (they are not equations) by another, as long as both limits exist, and the limit in the denominator is not equal to zero.

However, the questions that Ray raises make it difficult to understand exactly what the problem is. There's a difference between f(x) = 0 vs. f(a) = 0, and between f'(x) = 6 vs. f'(a) = 6. In the equation f(x) = 0, the assumption is that f is 0 at each value of x. In the equation f(a) = 0, the assumption is only that at x = a, the function value is 0. And similar for f'(x) vs. f'(a).

Is this the exact problem statement?

If f(x) = 0 and f'(x) = 6 find lim h -> 0 (f(x+h)/2h).

 

[Jamesaa1234]

You can divide one limit expression (they are not equations) by another, as long as both limits exist, and the limit in the denominator is not equal to zero.

However, the questions that Ray raises make it difficult to understand exactly what the problem is. There's a difference between f(x) = 0 vs. f(a) = 0, and between f'(x) = 6 vs. f'(a) = 6. In the equation f(x) = 0, the assumption is that f is 0 at each value of x. In the equation f(a) = 0, the assumption is only that at x = a, the function value is 0. And similar for f'(x) vs. f'(a).

Is this the exact problem statement?

would it be incorrect if i do not include the limit sign? Isnt first principle same as y2-y1/x2-x1.

 

[Mark44]

would it be incorrect if i do not include the limit sign?

No.

Isnt the limit same as y2-y1/x2-x1.

No. In the limit, the denominator is approaching zero.

Also, you need to use parentheses here. You wrote y2-y1/x2-x1, which is $y2 - \frac{y1}{x2} - x1$, surely not what you meant.

 

[Jamesaa1234]

No.

No. In the limit, the denominator is approaching zero.

Also, you need to use parentheses here. You wrote y2-y1/x2-x1, which is $y2 - \frac{y1}{x2} - x1$, surely not what you meant.

Okay just to clarify: my teacher took off marks and said you cannot find the ratio of two limits, is she incorrect?
Also could you check my work on the original post and let me know if there are any errors. Thank you.

 

[Mark44]

Okay just to clarify: my teacher took off marks and said you cannot find the ratio of two limits, is she incorrect?

If this is exactly what she said, yes, that's incorrect. It's a very well known theorem that if $\lim_{x \to a} f(x) = L$, and $\lim_{x \to a} g(x) = M \ne 0$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac L M$.

Also could you check my work on the original post and let me know if there are any errors. Thank you.

lim h-> 0 (2(f(a+h)/2h) = 6
lim h-> 0 (f(a+h)/2h) = 3

Can you show a bit more work here? What's your justification for the first limit?

 

[Jamesaa1234]

If this is exactly what she said, yes, that's incorrect. It's a very well known theorem that if $\lim_{x \to a} f(x) = L$, and $\lim_{x \to a} g(x) = M \ne 0$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac L M$.Can you show a bit more work here? What's your justification for the first limit?

Ik that the ratio between both limits is 2.

So i multiplied the two by the first limit and made it equal to the limit of f'(a) h-> 0 (2(f(a+h)/2h) = 6
then divided both sides by 2 to find lim for the original question.

Also, what is the theorem mentioned above called?

 

[Mark44]

Ik that the ratio between both limits is 2.

So i multiplied the two by the first limit and made it equal to the limit of f'(a) h-> 0 (2(f(a+h)/2h) = 6

What's your justification for saying that $\lim_{h \to 0} \frac{2f(a + h)}{2h} = 6$? That's not how f'(a) is defined.

then divided both sides by 2 to find lim for the original question.

Also, what is the theorem mentioned above called?

I don't know that it has a name. It's one of the properties of limits. There are similar properties for the limit of a sum of functions, product of functions, scalar multiple of a function, and quotient of two functions.

 

[Jamesaa1234]

What's your justification for saying that $\lim_{h \to 0} \frac{2f(a + h)}{2h} = 6$? That's not how f'(a) is defined.

f'(a) = 6
f(a) = 0
so (f(a+h) - f(a))/h = 6
f(a+h)/h = 6
Then i found the ratio between f(a+h)/h and f(a+h)/2h
The answer is 2. Which means 2(f(a+h)/2h) = f(a+h)/h
Therefore: 2(f(a+h)/2h) = 6 (value of f'(a))
and f(a+h)/2h = 3 (dividing both sides by 2)

 

[Mark44]

f'(a) = 6
f(a) = 0
so (f(a+h) - f(a))/h = 6

That's not exactly what I was looking for, but it's close.
Actually $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h} = 6$, so $\lim_{h \to 0} \frac{f(a + h)}{h} = 6$, since f(a) = 0.
Then $\lim_{h \to 0} \frac{f(a + h)}{2h} = \frac 1 2 \lim_{h \to 0} \frac{f(a + h)}{h} = 3$.

f(a+h)/h = 6
Then i found the ratio between f(a+h)/h and f(a+h)/2h
The answer is 2. Which means 2(f(a+h)/2h) = f(a+h)/h
Therefore: 2(f(a+h)/2h) = 6 (value of f'(a))
and f(a+h)/2h = 3 (dividing both sides by 2)