The limit at infinity of the function (3n+5)/(2n+7) is definitively calculated as 3/2. To prove this using an N epsilon approach, one must find a value N such that the absolute difference |(3n+5)/(2n+7) - 3/2| is less than ε for all n greater than N. This involves manipulating the expression to show that 2/(2n + 7) < ε when n > N, thereby establishing the limit rigorously.
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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?