## Rizags one year ago If f is differentiable at 1, and the limit of f(1+h)/h as h->0 is 5, what is f(1)?

1. Rizags

@jim_thompson5910

2. jim_thompson5910

 f is differentiable at 1 so $\Large \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$ exists and it is defined based on the limit definition of the derivative

3. Rizags

so $f'(1) = 5-\lim_{h \rightarrow 0}\frac{f(1)}{h}$

4. jim_thompson5910

$\Large \lim_{h\to 0}\frac{f(1+h)}{h}=5$ $\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-0}}}{h}=5$ $\Large \lim_{h\to 0}\frac{f(1+h){\LARGE \color{red}{-f(1)}}}{h}=5$ so we see that $$\Large f(1) = 0$$

5. Rizags

so my step would be invalid?

6. jim_thompson5910

it's valid, I just don't see where to go with it

7. Rizags

oh ok. And you knew to put in that f(1) as 0 simply because the equation resembled the derivative limit?

8. jim_thompson5910

I saw it match up with the limit definition. Well almost match up. It was just missing the -f(1) part

9. Rizags

so f(x) = 0 and f'(x) = 5?

10. jim_thompson5910

f ' (1) = 5 and f(1) = 0 we can't say anything about f(x) or f ' (x) in general

11. Rizags

ok, thank you

12. jim_thompson5910

IF it was $\Large \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=5$ then we can say f ' (x) = 5

13. Rizags

ok got it