x-> + infinity (x-sqrt(x^2-3x))
for this i got 3, because after it is rationalized i got 3x/x= 3
x-> - infinity (3x^2+4x+2)/(5-x+x^3)
i got 3/x , and zero as the final answer
are they right. thank you.
Help with this please?
The first one should be 3/2, we have:
sqrt(x^2-3x) =
x sqrt(1 - 3/x) =
x [1 - 1/2 (3/x) + O(1/x)^2] =
x - 3/2 + O(1/x)
sorry what is O, i don't understand
To find the limit of a function as x approaches positive or negative infinity, we need to examine the highest power of x in the numerator and denominator.
For the first limit, x -> + infinity (x - sqrt(x^2 - 3x)), notice that both the numerator and denominator contain the term x.
To determine the behavior of x - sqrt(x^2 - 3x) as x approaches positive infinity, we can simplify the expression:
x - sqrt(x^2 - 3x) = x - sqrt(x * (x - 3))
= x - sqrt(x) * sqrt(x - 3)
= x - sqrt(x) * sqrt(x) * sqrt(1 - (3/x))
Since x is approaching positive infinity, we can see that sqrt(1 - (3/x)) approaches sqrt(1 - 0) = 1.
Therefore, the expression simplifies to:
x - sqrt(x) * sqrt(x) = x - x = 0
So, as x approaches positive infinity, the limit of (x - sqrt(x^2 - 3x)) is 0.
Now, let's consider the second limit:
x -> - infinity (3x^2 + 4x + 2) / (5 - x + x^3).
In this case, the highest power of x is x^3 in the denominator.
To determine the behavior of (3x^2 + 4x + 2) / (5 - x + x^3) as x approaches negative infinity, we can divide every term in the numerator and denominator by x^3:
(3x^2 / x^3 + 4x / x^3 + 2 / x^3) / (5 / x^3 - x / x^3 + x^3 / x^3)
= (3/x + 4/x^2 + 2/x^3) / (5/x^3 - 1/x^2 + 1)
As x approaches negative infinity, all the terms with positive powers of x (3/x, 4/x^2, 5/x^3) tend to zero. The term 1/x^2 also approaches zero.
Therefore, the expression simplifies to:
(0 + 0 + 2/x^3) / (0 - 0 + 1) = 2/x^3
As x approaches negative infinity, the limit of (3x^2 + 4x + 2) / (5 - x + x^3) is 0.