Suppose $f(x)$ is differentiable at $x=1$ and ${\lim }_{h\to 0}\frac{f(1+h)}{h}=5$, then $f^{\prime }(1)$ equals:

Since $f(x)$ is differentiable at $x=1$, it must be continuous at $x=1$.For the limit ${\lim }_{h\to 0}\frac{f(1+h)}{h}$ to exist and be finite (5), the numerator must approach zero as $h\to 0$.${\lim }_{h\to 0}f(1+h)=0⟹f(1)=0$.By the definition of derivative:
$$f^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1+h)-f(1)}{h}$$
Substituting $f(1)=0$:
$$f^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1+h)}{h}=5$$