Question

предел функции как решить?

Answer 1

$\large \\ \lim_{x\rightarrow {-1}}{x^3+1 \over x^2-1}=\lim_{x\rightarrow -1}{x^3+1 \over (x-1)(x+1)}=\lim_{x\rightarrow -1}{(x+1)(x^2-x+1) \over (x-1)(x+1)}=\lim_{x\rightarrow -1}{x^2-x+1 \over x-1}=-{3\over2}\\\\ \lim_{x\rightarrow 3}{x^2-2x-3\over x^2-9}=\lim_{x\rightarrow 3}{(x^2-6x+9)+(-12+4x)\over (x-3)(x+3)}=\lim_{x\rightarrow 3}{(x-3)+(4x-12)\over (x-3)(x+3)}=\lim_{x\rightarrow 3}{(x-3)^2+4(x-3)\over (x-3)(x+3)}=\lim_{x\rightarrow 3}{x-3\over x+3}+4\lim_{x\rightarrow 3}{1\over x+3}=0+{4\over6}={2\over3}\\\\ \lim_{x\rightarrow \infty}{4x^2\over x^2-1}=\begin{pmatrix} {\infty\over \infty}\Rightarrow \div x^2 \end{pmatrix}=4\lim_{x\rightarrow \infty}{1\over 1-{1\over x^2}}=4$