Question

Lim h->0 (1 +h)² -1² / h

Answer 1

$\lim_{h\rightarrow0}\frac{(1+h)^2-1}{h}=\\\\\lim_{h\rightarrow0}\frac{[(1+h)+1][(1+h)-1]}{h}=\\\\\lim_{h\rightarrow0}\frac{(2+h)h}{h}=\\\\\lim_{h\rightarrow0}(2+h)=\\\\2+0=\\\\\boxed{2}$