[tex]\displaystyle\lim_{h\to0}\frac{(1+h)^9-1}h[/tex]
Write [tex]1=1^9[/tex], and recall that for a differentiable function [tex]f(x)[/tex], the derivative at a point [tex]x=a[/tex] is given by
[tex]f'(a)=\displaystyle\lim_{h\to0}\frac{f(a+h)-f(a)}h[/tex]
which would suggest that for this limit, [tex]f(x)=x^9[/tex] and [tex]a=1[/tex]. We have [tex]f'(x)=9x^8[/tex], and so the value of the limit would be [tex]9(1)^8=9[/tex].
