The discussion centers around finding the limit at infinity of the expression (3n+5)/(2n+7) and exploring the application of delta-epsilon and N-epsilon proofs in this context. Participants engage in mathematical reasoning related to limits and proofs.
Participants generally agree on the limit being $\frac{3}{2}$, but there is disagreement on the appropriate proof technique to use, with multiple approaches being discussed without a clear consensus on which is preferable.
Some participants express uncertainty about the steps involved in the proofs, and there are unresolved questions regarding the derivation of certain expressions related to the limit.
Instead of a delta epsilon proof, you need to use an N epsilon proof. In other words, given $\varepsilon>0$, you need to find $N$ such that $\Bigl|\frac{3n+5}{2n+7}-\frac{3}{2}\Bigr| < \varepsilon$ whenever $n > N$.dwsmith said:$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$
How does one use a delta epsilon proof for a limit at infinity?
Opalg said:Instead of a delta epsilon proof, you need to use an N epsilon proof. In other words, given $\varepsilon>0$, you need to find $N$ such that $\Bigl|\frac{3n+5}{2n+7}-\frac{3}{2}\Bigr| < \varepsilon$ whenever $n > N$.
dwsmith said:$\lim\limits_{n\to\infty}\frac{3n+5}{2n+7}=\frac{3}{2}$
How does one use a delta epsilon proof for a limit at infinity?
That's not what I got for \left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|. How did you get that?dwsmith said:Whenever $n > N$
$$
\frac{2}{2n + 7} < \epsilon
$$
Like this?
HallsofIvy said:That's not what I got for \left|\frac{3n+5}{2n+ 7}- \frac{3}{2}\right|. How did you get that?