# Solve lim_{xto1}frac{(x^{4}-4)}{(x^{2}-3x+2)}

Question
Limits and continuity
Solve $$\lim_{x\to1}\frac{(x^{4}-4)}{(x^{2}-3x+2)} ## Answers (1) 2021-02-23 If \(\lim_{x\to a-}f(x)\neq \lim_{x\to a+}f(x)$$ then the limit does not exist
$$\implies \lim_{x\to 1+}\frac{(x^{4}-4)}{(x^{2}-3x+2)}= \lim_{x\to 1+}((x^{4}-4)(x^{2}-3x+2)^{1})$$
$$= \lim_{x\to 1+}(x^{4}-4)\times\lim_{x\to 1+}((x^{2}-3x+2)^{-1})$$
$$=(-3)(-\infty)$$
$$=\infty$$
$$\implies \lim_{x\to 1-}\frac{(x^{4}-4)}{(x^{2}-3x+2)}= \lim_{x\to 1-}((x^{4}-4)(x^{2}-3x+2)^1)$$ $$= \lim_{x\to 1-}(x^{4}-4)\times \lim_{x\to 1-}((x^{2}-3x+2)^{-1})$$ Conclution: the \limit is diverges

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