The limit lim(x→0^+)(lnx/x) approaches -∞, as ln(x) approaches -∞ when x approaches 0 from the right. The expression cannot be simplified using l'Hospital's Rule because the conditions for its application are not met. Specifically, the limit involves a product of terms where one approaches infinity and the other approaches negative infinity, leading to an indeterminate form. Therefore, the correct evaluation of the limit is achieved by recognizing the behavior of ln(x) and the reciprocal of x as x approaches 0.
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It means that x approaches zero from the right (positive values close to zero).phillyolly said:Homework Statement
lim(x→0^+)(lnx/x)
The Attempt at a Solution
First, what does x->0+ mean? Is it positive infinity?
Look at the graph of y = ln(x). The domain is {x | x > 0}. As x gets closer to zero, y gets more and more negative.phillyolly said:Why lim(x→0^+) (lnx)=-∞?
What you have above is mostly incorrect. 1/0 is not a number, so you can't use it in calculations. \lim_{x \to 0^+} \frac{ln(x)}{x} = \lim_{x \to 0^+} \frac{1}{x}ln(x)phillyolly said:Why the answer is
= -∞ * (1/0+)
= -∞* ∞
= -∞?
Read the fine print in L'Hopital's Rule. There are certain conditions that must be satisfied before you can use it.phillyolly said:Why cannot I use l'Hospital Rule?