Another one I'm not sure about is the limit as x approaches infinity of (x^2)/lnx
I took the derivative of the top and bottom and got 2x/(1/x) which is the same thing as 2x^2. Can I keep going or do I just say that the limit does not exist?
you can keep going, but if you do, you're in trouble.
2x^2 = 2x^2/1
using the Rule, you'd have
4x/0 --> oo
So, the limit is as you found it, 2x^2 --> oo
To find the limit as x approaches infinity of the given function, (x^2)/lnx, you need to consider the behavior of the numerator and denominator separately.
First, let's analyze the numerator, x^2. As x approaches infinity, x^2 also approaches infinity because the square of any positive number grows without bound as the number itself increases. So, we can conclude that the numerator approaches infinity as x approaches infinity.
Next, let's examine the denominator, lnx. As x approaches infinity, the natural logarithm of x, lnx, also increases without bound but at a slower rate than x^2. This implies that the denominator approaches infinity at a slower pace than the numerator.
To determine the overall behavior of the function as x approaches infinity, you need to compare the growth rates of the numerator and denominator. In this case, since the numerator grows faster, the function approaches infinity as x approaches infinity.
Therefore, the limit as x approaches infinity of (x^2)/lnx is infinity.