1. ## Limits

Suppose that a function f is differentiable at x = 1 and
$\\lim_{h \to 0} \frac{f(1 + h)}{h} = 5$

Find f(1) and f '(1).

So, dy/dx at x = 1 is 5.

So the equation of the tangent line is y - f(1) = 5(x - 1).

But how can I find f(1)?

2. ## Re: Limits

Is my approach right?

I think if it is differentiable at x = 1, then it is also continuous at x = 1.

Then...

Any idea would be appreciable.

3. ## Re: Limits

This is what I think:
In general the definition of the derivative is:
lim_(h->0) (f(x+h)-f(x))/h = df(x)/dx
You have given a function which is differentiable at x=1 so:
lim_(h->0) (f(1+h)-f(1))/h = f'(1)
If you compare this with the given limit: f(1) = ...