# Evaluate $\lim\limits_{n \to \infty}\frac{sgn(n^2-3n+2)}{e^{n+1}}$

Solve : $$\lim\limits_{n \to \infty}\frac{sgn(n^2-3n+2)}{e^{n+1}}$$ Now, i started by applying the quotent rule $$\frac{\lim\limits_{n \to \infty} sgn(n^2-3n+2)}{\lim\limits_{n \to \infty}e^{n+1}}$$. The second limit gives out infinity . I do have a problem with the first limit : $$sgn(\lim\limits_{n \to \infty}(n^2-3n+2))$$ and then $$sgn(\infty)$$ which i am not sure how to evaluate . Wolfram says the answer is 0 .I just want an explanation for the first limit ( the one with the sign function ).

• for the signum part, you have to know, that this part is bounded by $1$. – Fakemistake Mar 17 at 10:41

$$0\xleftarrow[n\to\infty]{}\frac{-1}{e^{n+1}}\le\frac{\text{sgn}(n^2-3n+2)}{e^{n+1}}\le\frac1{e^{n+1}}\xrightarrow[n\to\infty]{}0$$
• Do you think we can take $(-1)^n$ instead of that $\text{sgn}$? – mrs Mar 17 at 10:51
When $$n \to \infty, n^2-3n+2 \sim n^2,~ so ~ sgn(n^2-3n+2)=+1$$ So the required limit is $$L=\lim_{n \to \infty} \frac{sgn(n^2-3n+2)}{e^{n+1}}=\frac{1}{\infty}=0$$