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Solve : $$\lim\limits_{n \to \infty}\frac{\mathrm{sgn}(n^2-3n+2)}{e^{n+1}}$$ Now, i started by applying the quotent rule $$\frac{\lim\limits_{n \to \infty} \mathrm{sgn}(n^2-3n+2)}{\lim\limits_{n \to \infty}e^{n+1}}$$The second limit gives out infinity . I do have a problem with the first limit : $\mathrm{sgn}(\lim\limits_{n \to \infty}(n^2-3n+2))$ and then $\mathrm{sgn}(\infty)$ 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 ).

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    $\begingroup$ for the signum part, you have to know, that this part is bounded by $1$. $\endgroup$ Mar 17, 2020 at 10:41

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Try the squeeze theorem:

$$0\xleftarrow[n\to\infty]{}\frac{-1}{e^{n+1}}\le\frac{\text{sgn}(n^2-3n+2)}{e^{n+1}}\le\frac1{e^{n+1}}\xrightarrow[n\to\infty]{}0$$

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  • $\begingroup$ Do you think we can take $(-1)^n$ instead of that $\text{sgn}$? $\endgroup$
    – Mikasa
    Mar 17, 2020 at 10:51
  • $\begingroup$ @mrs No, In each side that's exactly the sigh. Now, if you don't want to use the squeeze theorem but, say, the theorem that says that bounded times converging to zero is converging to zero then the answer is yes. $\endgroup$
    – DonAntonio
    Mar 17, 2020 at 12:26
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    $\begingroup$ Thanks my dear old friend! $\endgroup$
    – Mikasa
    Mar 17, 2020 at 17:46
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When $n \to \infty, n^2-3n+2 \sim n^2,~ so ~ \mathrm{sgn}(n^2-3n+2)=+1$ So the required limit is $$L=\lim_{n \to \infty} \frac{\mathrm{sgn}(n^2-3n+2)}{e^{n+1}}=\frac{1}{\infty}=0$$

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