Recent content by 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)

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?

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.

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

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.

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

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...