Find the lim x→∞ of (ln(x))^2)/x using L'hospital's rule.
I got - infinity when I did this, but I think I was looking at the numerator wrong when I took the derivative.
I got the answer, never mind!
To find the limit of the expression (ln(x))^2/x as x approaches infinity using L'Hôpital's Rule, we can differentiate the numerator and denominator separately until we reach a determinate form.
Let's start by taking the derivative of the numerator, (ln(x))^2, with respect to x. We can apply the chain rule here:
d/dx [(ln(x))^2] = 2(ln(x))(1/x) * 1/x [Using the chain rule and the power rule for derivatives]
Simplifying this further:
d/dx [(ln(x))^2] = 2(ln(x))/x^2
Now, let's calculate the derivative of the denominator, x, with respect to x:
d/dx (x) = 1
Now we have the derivatives of both the numerator and the denominator. Let's rewrite the original expression using these derivatives:
lim(x→∞) [(2(ln(x))/x^2) / 1]
Since x approaches infinity, we can see that the term 1/x^2 in the numerator will approach zero, and ln(x) will continue to grow as x becomes larger.
This means that the limit as x approaches infinity is:
lim(x→∞) [(2(ln(x))/x^2) / 1] = 2 * (∞) / (∞^2) = 2/∞ = 0
Therefore, the limit of (ln(x))^2/x as x approaches infinity is 0.