Question

vê se da p enxergar ae.. 

1264cda1ee6e740c0201044ca99d7a33.docx

Answer 1

$\lim_{x\rightarrow0}f(x)=\\\\\lim_{x\rightarrow0}\frac{(2+x)^2-4}{x}=\\\\\\\lim_{x\rightarrow0}\frac{(2)^2+2\cdot(2)\cdot(x)+(x)^2-4}{x}=\\\\\\\lim_{x\rightarrow0}\frac{4+4x+x^2-4}{x}=\\\\\\\lim_{x\rightarrow0}\frac{x^2+4x}{x}=\\\\\\\lim_{x\rightarrow0}\frac{x(x+4)}{x}=\\\\\\\lim_{x\rightarrow0}x+4=\\\\0+4=\\\\\boxed{4}$