limits to infinity

Question

lim x-> infinity (4-2x^2)/(2e^x)

lim x-> infinity (sqrt3x^6-x^3)/(7x^3+1)

for this one I got DNE? i first factored the top to absolute value x^3(3x^3-1) all over 7x^3+1. then set two equations of the positive and negative of the factored form. I got 1/7 and -1/7 after comparing the degrees and ratioed them.

does the constant always cancel when limits go to negative or positive infinity? is 1/x always 0 when it goes to positive infinity?

Answer 1

Have you learned l'Hopital's Rule? It states that given an f(x)/g(x) where both f(x) and g(x) approach 0 or +/- infinity as x goes to some value, you can take the limit of f'(x)/g'(x) and that limit equals the original limit..

If f(x) or g(x) go to 0 or infinity while the other reaches a finite limit, then the fraction is 0 or +/- ∞ (Therefore 1/x goes to 0 as x goes to ∞ and goes to ∞ as x approaches 0 from the right)

Lim as x→∞ of (4-x²)/2e^x fits the l'Hopital criterion of -∞/∞

derivative leads to -2x/2e^x (still same result)

derivative again leads to -2/2e^x which goes to 0 (note that any polynomial to any order will not go to infinity as fast as e^x and the fraction would always go to 0)

The 2nd problem is not clear what it is. You are correct that when you have a limit as x goes to infinity of a ratio of polynomials, the leading terms are the only thing that will remain after the limiting process. If the order of the leading terms are equivalent, then the ratio of the coefficients is the limit. The formal way to do this is to divide top and bottom by the highest order power of x. All of the terms except the leading term will go to 0.

If your problem is sqrt(3x⁶ -x³)/(7x³+1) divide top and bottom by x³

sqrt(3 - 1/x³)/(7+1/x³) = sqrt(3)/7 in the limit as x goes to infinity

Again, this is the formal way. You can see the limit by observing that we need only consider the leading terms sqrt(3)sqrt(x⁶)/7x³ with the x³ cancelling

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Answer 2

Okay, so for the first limit, you're likely using l'Hopital's rule as you've got an infinitely negative value over an infinite value. So you'd take the derivative of each to get it down to a simpler form.

= lim_x→∞ (4-2x²) / (2e^x) {Infinity over infinity}

= lim_x→∞ (-4x) / (2e^x) {Same scenario}

= lim_x→∞ (-4) / (2e^x) {Here the top isn't infinity, so we're good}

This process gets us a finite value over infinity, so the resulting limit equals zero.

Then for the next question has what is reasonably close to x³ in both the numerator and denominator. So just divide the top and bottom by that; for the square root, you'll insert it as though you were dividing by x⁶.

= lim_x→∞ √[(3x⁶-x³)/x⁶] / (7x³+1)/x³

= lim_x→∞ √(3-x^-3) / (7+x^-3)

= ^√3/₇

This is because those x^-3 go towards 0 as x goes towards infinity. So the answer is ^√3/₇.

Answer 3

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