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Find the limits.$$\lim _{x \rightarrow 0^{+}} \frac{1-\ln x}{e^{1 / x}}$$

$$\text { The limit is equal to } 0 .$$

Calculus 1 / AB

Chapter 3

TOPICS IN DIFFERENTIATION

Section 6

L Hopital's Rule; Indeterminate Forms

Functions

Limits

Derivatives

Differentiation

Continuous Functions

Applications of the Derivative

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here we have the limits as X goes to zero from the positive side of one minus natural Log X divided by E to the one over X. Now, if we try to do this limit notice in the numerator, we're going to get a positive infinity because the natural log by itself as a tends to zero from the right hand side equals negative infinity. So the numerator equals positive affinity the denominator We get a positive infinity. So we've infinity over infinity. That's an indeterminant form and we can go ahead and use low Patel's rule. We're gonna write in hell h of this equal sign indicating that that's what we do for this next step. Okay, so the derivative of the numerator is negative. One over X, the derivative the denominator. We're gonna have to do a chain rule. Let's see, So we get a negative X to the negative, too. Then we saw this and then this next step is simplifying. It simplifies here we get okay. And now this is a limit that weaken do because of the the numerator goes to zero. The denominator goes to infinity and that's it

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