# Finding the limit of $\lim_{k \rightarrow \infty} \left(\frac{2^k + 1}{2^{k-1} + 3}\right)$

$$\lim_{k \rightarrow \infty} \left(\frac{2^k + 1}{2^{k-1} + 3}\right)$$

I'm trying to prove that the limit of the sequence is $2$ using the squeeze theorem, but with no success. Thanks

• Is $k$ a constant? If so, since the fraction in the parenthesis does not depend on $n$, the limit can only be $0$, $1$ or $\infty$, never $2$ Commented Nov 16, 2013 at 12:43
• Sorry , I miss wrote the sequence , fixed it. Commented Nov 16, 2013 at 12:46
• Try multiplying both numerator and denominator by $2^{-(k-1)}$. Commented Nov 16, 2013 at 12:48
• I cant see how it helps can you elaborate ?. Commented Nov 16, 2013 at 12:52
• Then you did the multiplication wrong. You should get $\frac{2+1/2^{k-1}}{1+3/2^{k-1}}$. Commented Nov 16, 2013 at 12:55

HINT: Multiply the fraction by $1$ in the carefully chosen disguise
$$\frac{1/2^{k-1}}{1/2^{k-1}}\;.$$
since $$2^n + a \sim 2^n$$ $\forall n, a$ we have
$$\frac{2^{k} + 1}{2^{k-1} + 3} = \frac{2^k}{2^{k-1}} = 2$$