Question

My Calc is a little rusty:
Why is lim k-> infinity (k+1)/k =1?
and lim k-> infinity (2^(k+1))/2k = 2?

Answer 1

same as a horizontal asymptote
the degrees are the same, the limit is the ratio of the leading coefficients

Answer 2

trying plugging in bigger and bigger numbers into your function for k....
take the derivative of the bottom and top (1/1)=1

Answer 3

for the first one anyways

Answer 4

For the second one, did you mean to type:
lim k-> infinity (2^(k+1))/2^k = 2?

Answer 5

yes

Answer 6

Imagine k very large, like 10^6, and take the ratio, which becomes almost exactly 1

for the second try logs
(k+1) log 2 -log 2 - log k and let k get very large,

leaving k log 2

taking anti logs we have 2^k , which may be what you were looking for.

Answer 7

For the second one I think maybe you're just forgetting your rules of exponents yes?\[\Large\rm 2^{k+1}=2^k\cdot 2^1\]So that second limit becomes:\[\Large\rm \frac{2\cdot 2^k}{2^k}\]And then before taking your limit, you have a nice cancellation.

Answer 8

oh yes I did. Thanks so much!

Answer 9

The denominator was 2^k ??? Oh, well.

Answer 10

yeah sorry about that typo