Lim y→∞ 2-y / √7+6y^2?

Question

7+6y^2 is all under the square root

Answer 1

Hi,

lim x-->∞ 2-x / √(7+6x²)

There are a couple of ways of doing the problem. One is by dividing both the numerator and denominator by x:

2/x-x/x / √(7+6x²)/x = 2/x- 1 / √(7/x²+6)

lim x-->∞ 2/x- 1 / √(7/x²+6) = -1/√(6

Or, we could square both the numerator and denominator, keeping in mind that we have to keep track of the sign, and remembering that this operation introduces an extraneous root. When we do this, we get

lim x-->∞ (x² - 4·x + 4) / (6·x² + 7)

Dividing both the numerator and denominator by x² gives

lim x-->∞ (x² / x² - 4·x / x² + 4 / x²) / (6·x² / x²+ 7 / x²)

= lim x-->∞ (1 - 4 / x + 4 / x²) / (6 + 7 / x²) = 1/6

But, we have to take the square root and append the sign, which gives us

-1/√(6

Answer 2

Answer: 0

Here's the reason why:
For any Rational Expression, P/Q (that is, P(x) and Q(x) are both polynominals), if DegQ > Deg P, then Limit as x goes to infinity P(x)/Q(x) = 0.

Edit: I just read your note that 7 + 6y^2 is all under the square root sign. In this case, Answer is NOT 0. It's going to be - 1/sqr(6).

Answer 3

-1/√6

Answer 4

As y becomes really large, the denominator approaches y*sqrt(6). This leaves you with (2-y)/y, or 2/(y*sqrt(6)) - y/(y*sqrt(6)). As y approaches infinity, this approaches -1/sqrt(6).

Answer 5

answer
=-1/[6^(1/2)]