find limt
lim x approaches infinity 1/(5x+4)

- anonymous

find limt
lim x approaches infinity 1/(5x+4)

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- anonymous

0

- anonymous

denomiator gets bigger, numerator stays one. think of
\[\frac{1}{1000000}=.000001\]

- anonymous

bigger the denominator gets, the smaller the fraction.

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## More answers

- anonymous

ok so anything that deals with infinity could possibly be zero

- anonymous

well it depends on the function, but it certainly could be 0 in the limit.

- anonymous

ok. i have another one simliar but mor complex

- anonymous

shoot

- anonymous

\[ \lim \rightarrow \infty (1- x-x^2)/(3x^2-4)\]

- anonymous

ok in this case you have a polynomial of degree 2 in the numerator, and a polynomial of degree 2 in the denominator. when the degrees are the same, as in this case, just take the ratio of the leading coefficients.

- anonymous

leading coefficient of the numerator is -1, and the leading coefficient of the denominator is 3 so the limit as x goes to infinity is -1/3

- anonymous

what could be easier?

- anonymous

ok..so basically you look into the coefficent of both numerator and denominator and computer it to become a -1/3...so the -4 in the denomintor and the numerator dont neccessary needs anything

- anonymous

not a thing. lets do a simple example:
\[lim_{x->\infty}\frac{x^3+9x}{2x^2+x^2}\]

- anonymous

ok

- anonymous

and lets let x = 100 which is not even that big

- anonymous

ok cool

- anonymous

the numerator is
\[100^3+9\times 100=1000900\]

- anonymous

the denominator is 2010000

- anonymous

the ratio is
\[\frac{1000900}{2010000}\]

- anonymous

right and the denomator would be 2010000....

- anonymous

which is certainly not exactly \[\frac{1}{2}\]

- anonymous

ok..

- anonymous

but you can see that it is close. you can also see that the 9x term and the x^2 term meant nothing as far as the magnitude of the numerator and denominator

- anonymous

there are way down there in the decimal places.

- anonymous

i need a lot of practice on this function

- anonymous

and that was just for 100. we are taking the limit as x -> infinity. imagine what it would look like if x was 10000000000

- anonymous

very very close to 1/2

- anonymous

this is not how it is explained in your book. i am just pointing out that the numbers show you that you will get closer and closer to 1/2

- anonymous

i see becuase you can basically pick any number in the infinity column and it would still be the coeffiencet

- anonymous

now if the degree of the denominator is bigger than the numerator, then the limit is 0

- anonymous

think of
\[\frac{10^3}{10^5}\]

- anonymous

numerator has degree 3, denom has degree 5 and this number is small. it is .01

- anonymous

it would be zero because the 10^5 and large

- anonymous

last possibility is that the degree of the numerator is higher. in this case the limit is infinty

- anonymous

yes you are right.

- anonymous

now think of degree of the numerator bigger. for example
\[\frac{10^5}{10^2}=1000\]

- anonymous

and that is just for 10! as x -> infinity this will get huge. so the rules are as follows (and very simple)
lim x-> infity p(x)/q(x)

- anonymous

if deg p > deg q, limit is infinity
if deg p < deg q limit is 0
if deg p = deg q ratio of leading coefficients

- anonymous

that is it!

- anonymous

ok no i see thanks again!!

- anonymous

welcome!

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