Find $\displaystyle \lim_{n \to \infty} \frac{2^{n+1}-3}{2^n+1}$ using the definition of a limit.
Attempt: For $\epsilon > 0$ take $N = \dfrac{\log{(1/\epsilon)}}{\log(2)}$. Then $n > N$ implies $\log(2^n) > \log(1/\epsilon)$ which implies $2^n > 1/\epsilon$ which gives $\dfrac{1}{2^n} < \epsilon$. But $\displaystyle \frac{1}{2^n} > \frac{1}{2^n+1} > \frac{-5}{2^{n}+1}$ so we have $\displaystyle \frac{-5}{2^{n}+1} < \epsilon$ therefore
$\bigg|\dfrac{2^{n+1}-3-2(2^{n}+1)}{2^{n}+1}\bigg| < \epsilon \implies \bigg|\dfrac{2^{n+1}-3}{2^n+1}-2\bigg| < \epsilon$. Therefore we have $\displaystyle \lim_{n \to \infty} \frac{2^{n+1}-3}{2^n+1} = 2.$
Is this correct? If not, how do you do it? And if it's how do ensure that $N = \dfrac{\log(1/\epsilon)}{\log(2)}$is an integer?
This is correct. To ensure that $N$ is an integer, use the ceiling function $$ N = \lceil \frac{-\log \epsilon}{\log 2} \rceil $$