Expand everything in the limit:
[tex]\displaystyle\lim_{h\to0}\frac{(-7+h)^2-49}h=\lim_{h\to0}\frac{(49-14h+h^2)-49}h=\lim_{h\to0}\frac{h^2-14h}h[/tex]
We have [tex]h[/tex] approaching 0, and in particular [tex]h\neq0[/tex], so we can cancel a factor in the numerator and denominator:
[tex]\displaystyle\lim_{h\to0}\frac{h^2-14h}h=\lim_{h\to0}(h-14)=\boxed{-14}[/tex]
Alternatively, if you already know about derivatives, consider the function [tex]f(x)=x^2[/tex], whose derivative is [tex]f'(x)=2x[/tex].
Using the limit definition, we have
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h=\lim_{h\to0}\frac{(x+h)^2-x^2}h[/tex]
which is exactly the original limit with [tex]x=-7[/tex]. The derivative is [tex]2x[/tex], so the value of the limit is, again, -14.
