You can recognize the limit as a derivative. Recall that
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
and more to the point,
[tex]f'(1)=\displaystyle\lim_{h\to0}\frac{f(1+h)-f(1)}h[/tex]
We have
[tex]f(x)=x^4\implies f'(x)=4x^3\implies f'(1)=4[/tex]
so the answer is B.
We don't actually have to invoke the definition of the derivative. Instead we can just use the definition of [tex]f(x)[/tex]:
[tex]\displaystyle\lim_{h\to0}\frac{(1+h)^4-1^4}h=\lim_{h\to0}\frac{1+4h+6h^2+4h^3+h^4-1}h=\lim_{h\to0}(4+6h+4h^2+h^3)[/tex]
which you can see also approaches 4.