If b>0 prove that lim_{n->infinity} ^{1}/1+nb=0 using the limit definition.
I can't seem to make any headway in choosing my delta to make progress in this exercise.
what we need to do is show that for ANY ε > 0 (even, or perhaps especially the very tiny ones), we can find SOME positive integer N, such that:
for ALL n > N:
$\displaystyle \left|\frac{1}{1+nb} - 0\right| < \epsilon$
note we're not using "delta" because n isn't tending to a certain definite real number (|x - ∞| < δ doesn't make any sense). instead, we're using the idea that "close to infinity" means "really big".
we expect that if ε is very small, N should be quite large.
now:
$\displaystyle \left|\frac{1}{1+nb} - 0\right| = \left|\frac{1}{1+nb}\right| = \frac{1}{1+nb}$ (since we can insist n > 0 (we are, after all, going "all the way positive to infinity") and we know that b > 0).
if we want:
$\displaystyle \frac{1}{1+nb} < \epsilon$, then we want:
$\displaystyle 1 < (1+nb)\epsilon$, so:
$\displaystyle \frac{1}{\epsilon} < 1 + nb \iff \frac{1}{\epsilon} - 1 < nb \iff \left(\frac{1}{b}\right)\left(\frac{1}{\epsilon} - 1\right) < n$
this suggests we pick $\displaystyle N > \left(\frac{1}{b}\right)\left(\frac{1}{\epsilon} - 1\right)$, or N = 1 (just to keep N positive).
so, if we choose:
$\displaystyle N = \max\left(1,\frac{\1-\epsilon}{b\epsilon}\right)$, we have for all n > N:
$\displaystyle \left|\frac{1}{1+nb}} - 0\right| = \frac{1}{1+nb} < \frac{1}{1+Nb} < \frac{1}{1 + \left(\frac{1-\epsilon}{b\epsilon}\right)b}$
$\displaystyle =\frac{1}{\frac{\epsilon + 1 - \epsilon}{\epsilon}} = \frac{1}{\frac{1}{\epsilon}} = \epsilon$, as desired.
let's see how this works for a specific ε, and a specific b (we'll actually find the N that works).
suppose b = 2, and ε = 0.1 or 1/10.
according to what we did above, the integer N we want is:
N > (1/2)(9/10)(10) = 9/2, so we should pick N = 5.
note that 1/(1 + 2*4) = 1/9 > 1/10, so N = 4 doesn't work.
now if n ≥ 5:
1/(1 + 2n) < 1/11 < 1/10.