[tex]\displaystyle\lim_{h\to0}\frac{(4+h)^5-1024}h=\lim_{h\to0}\frac{(4+h)^5-4^5}h[/tex]
Recall the definition of the derivative of a function [tex]f(x)[/tex] at a point [tex]x=c[/tex]:
[tex]f'(c)=\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)}h[/tex]
We can then see that [tex]f(c)=c^5[/tex], and by the power rule we have [tex]f'(c)=5c^4[/tex]. Then replacing [tex]c=4[/tex], we arrive at
[tex]\displaystyle\lim_{h\to0}\frac{(4+h)^5-4^5}h=5\times4^4=1280[/tex]
Alternatively, we could have expanded the binomial, giving
[tex]\dfrac{(4+h)^5-4^5}h=\dfrac{(4^5+5\times4^4h+10\times4^3h^2+10\times4^2h^3+5\times4h^4+h^5)-4^5}h[/tex]
[tex]=\dfrac{1280h+640h^2+160h^3+20h^4+h^5}h[/tex]
[tex]=1280+640h+160h^2+20h^3+h^4[/tex]
and so as [tex]h\to0[/tex] we're left with 1280, as expected.
