Demonstrate by definition the following limit

Question

Demonstrate by definition the following limit

∀ ε>0 Ǝ M>0 / x>M ,|a_x-L|< ε

$$\lim_{x \to \infty} \frac{x^2+4}{(x-1)(x+2)(x-3)} = 0$$

Answer 1

Let $$f(x) = \frac{x^2+4}{(x-1)(x+2)(x-3)}$$ We want to prove that $$\lim_{x \to \infty} f(x) = 0$$ by showing that for all $\epsilon > 0$, there exists $M > 0\;$such that $$x > M \implies |f(x)| < \epsilon$$ Fix $\epsilon > 0$, and let ${\displaystyle{M = \frac{2}{\epsilon}+8}}$. \begin{align*} \text{Then}\;\;&M = \frac{2}{\epsilon}+8\\[4pt] \implies\;&M-8=\frac{2}{\epsilon}\\[4pt] \implies\;&\frac{2}{M-8}=\epsilon\\[4pt] \end{align*} Then for any $x > M$, \begin{align*} |f(x)| &= \left|\frac{x^2+4}{(x-1)(x+2)(x-3)}\right|\\[4pt] &=\frac{x^2+4}{(x-1)(x+2)(x-3)}&&\text{[since $x > M > 8$]}\\[4pt] &<\frac{x^2 + 4x + 4}{(x-1)(x+2)(x-3)}\\[4pt] &=\frac{(x+2)^2}{(x-1)(x+2)(x-3)}\\[4pt] &=\frac{x+2}{(x-1)(x-3)}\\[4pt] &<\frac{x+2}{(x-3)^2}\\[4pt] &=\frac{x-3}{(x-3)^2}+\frac{5}{(x-3)^2}\\[4pt] &=\frac{1}{x-3}+\frac{5}{(x-3)^2}\\[4pt] &<\frac{1}{x-3}+\frac{x-3}{(x-3)^2}&&\text{[since $x > M > 8 \implies x-3 > 5$]}\\[4pt] &=\frac{1}{x-3}+\frac{1}{x-3}\\[4pt] &=\frac{2}{x-3}\\[4pt] &<\frac{2}{x-8}\\[4pt] &<\frac{2}{M-8}\\[4pt] &=\epsilon\\[4pt] \end{align*} as was to be shown.