Question

find the limit or show that it does not exist.

lim (x-4)/(x^2+7)
x--> infinity

Answer 1

degree of the denominator is larger than the degree of the numerator
so limit is 0

Answer 2

In general, if the degree of the polynomial on the top is less than that on the bottom, then the limit is 0. Try that with l'Hopital's Rule. Beaten!

Answer 3

this is identical to a pre calc question " find the horizontal asymptote if any" and the answer would be \(y=0\)

Answer 4

think of what would happen if you replaced \(x\) by \(1000000=10^6\) numerator would have 6 places, but the denominator would have 12!

Answer 5

ok. so determining the infinite limits are the same as finding the horizontal asymptote? And the vertical asymptotes are not related in this way?

Answer 6

good explanations, thank you

Answer 7

yes, it is the same

Answer 8

well, it is the same if you are looking for a limit as x goes to infinity
not for example as x goes to zero

Answer 9

and if there is a square root in the expression does the same rule apply?

Answer 10

so you can use the same rules you knew from whatever pre calc class you had

Answer 11

essentially
for example
\[\lim_{x\to \infty}\frac{5x}{\sqrt{6x^2+x+4}}=\frac{5}{\sqrt{6}}\]

Answer 12

ok, so the limit would still be 0 because the denominator has a larger degree than the numerator

Answer 13

thank you

Answer 14

in my example the limit was \(\frac{5}{\sqrt{6}}\) because \(\sqrt{6x^2}\) is like
\[\sqrt{6}x\] so the degrees are the same, both 1

Answer 15

ok, I was thinking you had to take out the entire expression out of the radical in order for them to be the same