Suppose f(x) is differentiable for all x and $\lim\limits_{h \rightarrow 0} \frac{1}{h} f(1+h)=5$, then f'(1) equals

5

It is given that f(x) is differentiable for all x. So, it is everywhere continuous. Consequently, f(1) is a finite quantity.

Now,

$f'(1) = \lim\limits_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

$\Rightarrow f'(1) =\lim\limits_{h \rightarrow 0} \frac{1}{h} f(1+h)-\lim\limits_{h \rightarrow 0} \frac{f(1)}{h}$

$\Rightarrow f'(1)=5-\lim\limits_{h \rightarrow 0} \frac{f(1)}{h}$

$\Rightarrow \lim\limits_{h \rightarrow 0} \frac{f(1)}{h}=5-f'(1)$

Clearly, RHS is a finite quantity as f'(1) exists. So, LHS must also be finite. This is possible only when f(1) = 0.

∴  0 = 5 - f'(1) ⇒ f'(1) = 5