Question

Find the limit:

Answer 1

\[\lim_{x \rightarrow \infty}(6x+1)/(3x+8)\]

Answer 2

do it with your eyeballs
degree of numerator is the same as the degree of the denominator, (they are both one) so take the ratio of the leading coefficients

Answer 3

in your case it is \(\frac{6}{3}=2\)

Answer 4

that easy huh? what about something with different degrees? like \[\lim_{x \rightarrow \infty}(3x ^{2}+1)/x\]

Answer 5

then you have to compare degrees:
higher numerator, limit is infinite ( or doesn't exist)
higher denominator, limit is zero

Answer 6

here's the rule:
if the degree on the top = degree on the bottom, then the limit is the ratio of the coefficients of the highest degree terms

if the degree on the top > degree on the bottom, limit DNE
if the degree on the top < degree on the bottom, limit is zero.

Answer 7

ok i'm starting to recall now. thank you both for your help.

Answer 8

think about if x was a power of ten, say \(x=1000\)
then in your example you would have
\[\frac{3\times 1000^2+1}{1000}\] which is
\[\frac{3000000+1}{1000}\approx3000\] already

Answer 9

whereas
\[\frac{6\times 1000+1}{3\times 1000+8}=\frac{6001}{3001}\approx 2\]

Answer 10

ok thank you @satellite73. great explanation

Answer 11

my pleasure