Question

Limits involving infinity:

Answer 1

2x^3 + 7 / x^3 - x^2 +x +7

Answer 2

the limit as x -> infinity?

Answer 3

divide the numerator and denomincator by the highest x term
so x^3

Answer 4

then find the limit
remember
lim x->infinity for 1/x = 0

Answer 5

limit x - positive infinity and lim x - negative infinity

Answer 6

i divided by x to get 7/x on numerator and denominator so it approaches zero. But idk what to do after that.

Answer 7

Factor out a x^3 from top and bottom:

2x^3 + 7/ x^3-x^2 + x +7 = x^3( 2+7/x^3) / x^3( 1 - 1/x + 1/x^2 + 7)

x^3 cancels out from the top and bottom so you are left with:

(2+(7/x^3)) /(1- (1/x) + (1/x^2) + (7/x^3))

now lets look at what happens to 1/x as x-> infinity (both positive and negative)

well if x is getting bigger and bigger, than 1/x is getting smaller and smaller. So the limit as x -> infinite of 1/x is just 0. Knowing this, anything in the form of 1/x, 1/x^2, and 1/x^3 is also going to 0 as x-> infinity (it does not matter if it is positive or negative infinity, since its heading towards zero).

so the above expression would look like this, after we take the limits of 7/x^3, 1/x, 1/x^2:

(2 + 0) / (1 - 0 + 0 +0) = 2/1 = 2

So the whole thing is going to 2, if x-> infinite (both positive and negative).

Answer 8

meh latex is annoying
\[\lim_{x \rightarrow \infty} \frac{2x^3 + 7}{ x^3 - x^2 +x +7}\]
\[\lim_{x \rightarrow \infty} \frac{2x^3 + 7}{ x^3 - x^2 +x +7}*1\]
\[\lim_{x \rightarrow \infty} \frac{2x^3 + 7}{ x^3 - x^2 +x +7}*\frac{\frac{1}{x^3}}{\frac{1}{x^3}}\]
\[\lim_{x \rightarrow \infty} \frac{\frac{2x^3}{x^3} + \frac{7}{x^3}}{ \frac{x^3}{x^3} - \frac{x^2}{x^3} +\frac{x}{x^3} +\frac{7}{x^3}}\]
now simplify

Answer 9

so what if it is 1/ x^3 - 4x +1? Thank you so much btw, helped me a lot with understanding my calc hw :)

Answer 10

you could do what i did, which was divide top and bottom by the highest x exponent

or you could just realize the denominator goes to infinity and thus the limit is equal to 0

assuming its going to infinity

Answer 11

OH ok... Thank you so much guys! I was scared I wouldn't get any answers because not a lot of people reply to calculus stuff haha

Answer 12

Yeah, an easier way to go is to notice that, for some polynomial \(P(x)\):
\[
P(x)=a_nx^n+...+a_0\\
P(x)\sim a_nx^n, \; x\to \infty\implies\\
\lim_{x \to \infty}f(P(x))=\lim_{x \to \infty}f(a_nx^n)
\]

Answer 13

so if the answer was 2/ 7/x^3 the limit would be approaching zero, however... the problem (after simplified) states 2/1. Therefore the limit is approaching 2... is that correct?

Answer 14

Yes, were it:
\[
\lim_{x\to \infty}\frac{\frac{2}{7}}{x^3}=0
\]But:
\[
\lim_{x \to \infty}\frac{2x^3+7}{x^3-x^2+x+7}=\lim_{x\to \infty}\frac{2x^3}{x^3}=\frac{2}{1}=2
\]Since the other terms become insignificant as \(x \to \infty\)