Using l'hopital's rule, find the limit as x approaches infinity of
(e^(6/x)-6x)^(X/2)
I know l'hopital's rule, but this is seeming to be nightmare. I just don't seem to get anywhere.
To find the limit as x approaches infinity of the given expression using L'Hôpital's rule, we need to take the derivative of both the numerator and the denominator until we reach a form that is easier to evaluate.
Let's begin by applying L'Hôpital's rule to the expression.
1. Take the derivative of the numerator:
d/dx (e^(6/x) - 6x)
= d/dx (e^(6/x)) - d/dx (6x)
The derivative of e^(6/x) can be found using the chain rule:
= e^(6/x) * d/dx (6/x) - 6
Simplifying further, we get:
= e^(6/x) * (-6/x^2) - 6
= -6e^(6/x)/x^2 - 6
2. Now take the derivative of the denominator:
d/dx (x/2)
= 1/2
So, the expression becomes:
[-6e^(6/x)/x^2 - 6] / (1/2)
To simplify it further, let's multiply the numerator and denominator by 2:
[-12e^(6/x)/x^2 - 12] / 1
Now, let's take the limit as x approaches infinity again by applying L'Hôpital's rule once more.
1. Take the derivative of the numerator:
d/dx [-12e^(6/x)/x^2 - 12]
= d/dx [-12e^(6/x)/x^2] - d/dx (12)
The derivative of -12e^(6/x)/x^2 can be found using the quotient rule:
= (-12/x^2) * e^(6/x) - (-12)(2x)/x^4
Simplifying further, we have:
= -12e^(6/x)/x^2 + 24/x^3
2. The derivative of the denominator is zero since it is a constant.
So, the expression becomes:
[-12e^(6/x)/x^2 + 24/x^3] / 1
To simplify further, we can factor out 1/x^3 from the numerator:
= (1/x^3)[-12e^(6/x)/x + 24] / 1
Now, take the limit as x approaches infinity:
= (1/x^3)[-12e^(6/x)/x + 24] / 1
= (-12/e^0)(0 + 24) / 1
= -12(24) / e^0
= -288.
Therefore, the limit as x approaches infinity of (e^(6/x) - 6x)^(x/2) is -288.