Can anyone solve this? Find the limit of 3 absolute value of x/x+2 as x approaches negative infinity.?

Question

Please explain your answer. Thanks!

That's 3 times the absolute value of x/x+2

Answer 1

Assuming the function is 3|x| / (x+2)....

When x < 0, |x| is the same as -x; so your function can be written as
lim 3(-x) / (x+2)
= lim -3x / (x+2)
= lim -3( x +2 - 2) / (x+2)
= lim -3(x+2)/(x+2) + lim (-3)(-2)/(x+2)
= lim (-3) + lim 6/(x+2)
= -3 + lim 6/(x+2)

As x -> -infinity, 6/(x+2) -> 0; so the answer is -3 .

(You can check this by using an Excel spreadsheet)

Answer 2

Hi, SpunkiiMonkii! We can only use L'Hopital's Rule for indeterminate forms such as 0/0 and ∞/∞ Fortunately, this expression can be written as the quotient of two monomials: x^3 e^(-x^2) = x^3 / e^(x^2) This is in the form ∞/∞, so we use L'Hopital's Rule and take the derivative with respect to x of both the numerator and denominator. The first derivative yields: 3x^2 / 2xe^(x^2) ... which is still in the indeterminate form ∞/∞. So we can apply L'Hopital's Rule again. But before we do that, let's think about what's happening: As we continue to take derivatives, the degree of the monomial expression in the numerator will continue to decrease, until we end up with a constant value. However, in the denominator, we can never get rid of the exponential factor containing x. So, the denominator will always tend toward infinity, and the limit is 0. To see this explicitly, note that once we carry out L'Hopital's Rule a few times, we end up with: 6 / (12xe^(x^2) + 8x^3 e^(x^2)) As x goes to infinity, the numerator remains constant, whereas the denominator goes to infinity. So, as we determined earlier, the limit really is 0.

Answer 3

First of all, take the expression "x/(x+2)". Put in facor x, you obtain (x(1))/(x(1+2/x)). Simplify the quotient, you have now 1/(1+2/x), which tend to 1 when x tend to negative infinite. So the 3 absolute value tend to 3*1=3.

Answer 4

lim(x--> -∞) 3|x/(x+2)|

3|x/(x+2)|

Multiply the inside of the abs. value by (1/x)/(1/x)
= 3|1/(1 + 2/x)|

lim(x--> -∞) 3|1/(1 + 2/x)|
= 3|1/(1 + 2/(-∞))|
= 3|1/(1 + 0)|
= 3|1|
= 3

Answer 5

3|x/(x+2|)
lim x--> - inf
This gives - inf/- inf so use L'Hospital's Rule getting
3|1/1|= 3