- $\lim_{(n) \rightarrow (\infty)} \frac{7n^2+9n-17}{4n^2-5n+6}= 7/4$
I got that $|\frac{71n-110}{16n^2-20n+24}| < $ $\epsilon$ , how do I continue from here ?
2026-03-28 06:10:23.1774678223
prove the limit using definition. $\lim_{(n) \rightarrow (\infty)} \frac{7n^2+9n-17}{4n^2-5n+6}= 7/4$
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It is, for any $n\ge 3,$ $$\left|\frac{7n^2+9n-17}{4n^2-5n+6}- \frac74\right|= \left|\frac{71n-110}{16n^2-20n+24}\right|<\frac{71n}{16n^2-20n}=\frac{71}{4(4n-5)}<\frac{71}{4(n-2)}.$$
Can you finish now?