Question

Answer 1

\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2-1\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(x^2-1\right)}{x^2+2x+4}=1\)