- #1
Mr Davis 97
- 1,462
- 44
I am trying to show that ##\displaystyle \lim \frac{1}{6n^2+1}=0##.
First, we have to find an N such that, given an ##\epsilon > 0##, we have that ##\frac{1}{6n^2+1} < \epsilon##. But in finding such an N, I get the inequality ##n> \sqrt{\frac{1}{6}(\frac{1}{\epsilon}-1)}##. But clearly with the square root we would have to have that ##\epsilon>1##, which doesn't make any sense since epsilon can be any positive real number. What am I doing wrong?
First, we have to find an N such that, given an ##\epsilon > 0##, we have that ##\frac{1}{6n^2+1} < \epsilon##. But in finding such an N, I get the inequality ##n> \sqrt{\frac{1}{6}(\frac{1}{\epsilon}-1)}##. But clearly with the square root we would have to have that ##\epsilon>1##, which doesn't make any sense since epsilon can be any positive real number. What am I doing wrong?