find limt lim x approaches infinity 1/(5x+4)
0
denomiator gets bigger, numerator stays one. think of \[\frac{1}{1000000}=.000001\]
bigger the denominator gets, the smaller the fraction.
ok so anything that deals with infinity could possibly be zero
well it depends on the function, but it certainly could be 0 in the limit.
ok. i have another one simliar but mor complex
shoot
\[ \lim \rightarrow \infty (1- x-x^2)/(3x^2-4)\]
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.
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
what could be easier?
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
not a thing. lets do a simple example: \[lim_{x->\infty}\frac{x^3+9x}{2x^2+x^2}\]
ok
and lets let x = 100 which is not even that big
ok cool
the numerator is \[100^3+9\times 100=1000900\]
the denominator is 2010000
the ratio is \[\frac{1000900}{2010000}\]
right and the denomator would be 2010000....
which is certainly not exactly \[\frac{1}{2}\]
ok..
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
there are way down there in the decimal places.
i need a lot of practice on this function
and that was just for 100. we are taking the limit as x -> infinity. imagine what it would look like if x was 10000000000
very very close to 1/2
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
i see becuase you can basically pick any number in the infinity column and it would still be the coeffiencet
now if the degree of the denominator is bigger than the numerator, then the limit is 0
think of \[\frac{10^3}{10^5}\]
numerator has degree 3, denom has degree 5 and this number is small. it is .01
it would be zero because the 10^5 and large
last possibility is that the degree of the numerator is higher. in this case the limit is infinty
yes you are right.
now think of degree of the numerator bigger. for example \[\frac{10^5}{10^2}=1000\]
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)
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
that is it!
ok no i see thanks again!!
welcome!
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