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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to \theta} \frac{\ln \sqrt{x}}{x^2}$

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in this case, what we're given is the natural log of X. Central log of the square root of X. Actually divided by X square. So we do that. If we plug in the value directly infinity, we end up getting infinity over infinity and this is not allowed. So what we're gonna do instead is um take the derivative of the top and the bottom. So something we can do in this case is Um substitute the square root of x. 40. So that allow this to be T here And then that'll make this T to the 4th. The reason why we prefer to do this is because it makes little tells you a little bit easier to apply when we plug in um infinity for X. That will give us T equals infinity. So that's infinity over infinity and we don't want that. So now we're going to take the derivative of the top and bottom, that'll give us one over T. And this will be uh four T cubed before T cube. Now, when we plug in, X equals infinity, or T equals infinity. That's going to end up giving us zero. Um this right here would be zero and the numerator, and then in the denominator we would get uh infinity, so zero over infinity is just going to be zero. So that would be our final answer.

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