Question

Limit Question

Answer 1

When taking the limit:

\[\lim_{n \rightarrow \infty}\frac{ 2n^2 }{ 4n^4 +4}\]

Why cant you use the exponent rule, where if the exponents are the same the limit = the coefficient of the numerator over coefficient of denominator? I see why this goes to zero but my inital instinct was to use the rule I mentioned and I got the question wrong, thinking the limit approached (1/2)

Answer 2

*** where if the exponents are the same ***

but the exponents are not the same

Answer 3

Sorry, I wrote the problem wrong

Answer 4

\[\lim_{n \rightarrow \infty}\frac{ 2^n }{ 4^n+4 }\]

Answer 5

your rule does not apply when the exponent is n

one easy way to do this is to divide top and bottom by 2^n

Answer 6

\[\frac{ 1 }{ 2^n \frac{ 4 }{ 2^n } }\] ?

Answer 7

that must not be right

Answer 8

4^n / 2^n is the same as (4/2)^n
or 2^n
so
\[ \frac{1}{2^n+ \frac{4}{2^n}} \]

Answer 9

as n-> infinity the bottom goes to infinity
so the fraction goes to zero

Answer 10

oh I see, I forgot it was addition and was thinking the 2^n would cancel out. I see how it goes to zero now. Thank you @phi

Answer 11

if we want to be pedantic, we would do
limit of the top divided by the limit of the bottom

of course the limit of the top remains 1
in the bottom, the limit of the sum is the sum of the limits
i.e. lim( 2^n) + lim(½^n)
the 2nd limit goes to zero
the first goest to infinity
so we get 1/infinity which is approaching zero

Answer 12

Pedantics help, Im suddenly having trouble with limits for some reason so ways to check my work are useful.