# Homework Help: Limit Question using first principles

1. Mar 4, 2016

### Jamesaa1234

1. The problem statement, all variables and given/known data
If f(a) = 0 and f'(a) = 6 find lim h -> 0 (f(a+h)/2h).

2. Relevant equations
lim h ->0 (f(a+h)-f(a))/h

3. 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?)

Last edited: Mar 4, 2016
2. Mar 4, 2016

### Ray Vickson

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.

3. Mar 4, 2016

### Jamesaa1234

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

4. Mar 4, 2016

### Ray Vickson

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.

5. Mar 4, 2016

### Jamesaa1234

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.

Last edited: Mar 4, 2016
6. Mar 4, 2016

### Staff: Mentor

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?

7. Mar 4, 2016

### Jamesaa1234

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

8. Mar 4, 2016

### Staff: Mentor

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.

9. Mar 4, 2016

### Jamesaa1234

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.

10. Mar 4, 2016

### Staff: Mentor

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 $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac L M$.
Can you show a bit more work here? What's your justification for the first limit?

11. Mar 4, 2016

### Jamesaa1234

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?

Last edited: Mar 4, 2016
12. Mar 4, 2016

### Staff: Mentor

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.
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.

13. Mar 5, 2016

### Jamesaa1234

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)

14. Mar 5, 2016

### Staff: Mentor

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 $\lim_{h \to 0} \frac{f(a + h)}{2h} = \frac 1 2 \lim_{h \to 0} \frac{f(a + h)}{h} = 3$.