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For $ f(x) = \frac{2}{x} - \frac{1}{\ln x} $ find…

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Problem 43 Medium Difficulty

(a) For $ f(x) = \frac{x}{\ln x} $ find each of the following limits.

(i) $ \displaystyle \lim_{x \to 0^+} f(x) $
(ii) $ \displaystyle \lim_{x \to 1^-} f(x) $
(iii) $ \displaystyle \lim_{x \to 1^+} f(x) $

(b) Use a table of values to estimate $ \displaystyle \lim_{x \to \infty} f(x) $.

(c) Use the information from parts (a) and (b) to make a rough sketch of the graph of $ f $.

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Video Transcript

and this activity were being given this function f of X equals X divided by the log of X. And for part a we're asked to find three limits. The first limit he is, what is the limit of this function? As the input? X approaches zero from the positive side. Okay, well, as X approaches zero, the numerator is approaching zero. So this is a small number. And as X approaches zero in the denominator, this is a negative larger number as the number gets smaller and smaller and smaller and is divided by a number that's getting bigger and bigger and bigger than the total value is approaching zero But it is approaching zero from the negative side. So the limit is going to be coming from the negative side and approaching zero as X approaches zero from the positive side, which means looking on a graph. Yeah. Right. Yeah. Yes. Mhm. That portion of the ground that's close to zero on the positive side. Mhm. We'll be approaching zero but from the negative sides, that part of the graph will look kind of like that. Then the second thing that were asked to consider is what is the limit of this function As X approaches one from the negative side? Well, again, if X is approaching negative effects is approaching one from the negative side, the limit of the numerator is going to be one. It's going to be a number that's just slightly smaller than one but it's getting closer and closer to once that's essentially constant. The log of one is zero. The log of a number that is less than one is a negative value that is getting closer and closer to zero. So if I have a constant divided by a very small number that's getting smaller and smaller and smaller and is negative then that is approaching a negative infinity. And the third limit that were asked to consider in part A is what is the limit of this function F has X approaches one from the positive side and Using the same logic, I've got a constant one that is being divided by a number that is getting smaller and smaller and smaller but that is positive and that is going to be a positive infinity. So on hoops. Mhm. Yeah. Yeah. My God. As the X is approaching one, I put my ass into it in the wrong spot. So The limit has x approaches zero is zero. It's coming from the negative side. The limit as X approaches one from the negative side is negative infinity. The limit as X approaches one from the positive side, his positive infinity. And then we're being asked to use a table of values in part B to find what is the limit of dysfunction F as X approaches infinity. Yeah. Yeah. Yeah. So if I calculate this function f with an input of 1000 to say mhm. That gives me 144. If I evaluate this function with an input of one million for example. Yeah, that gives me an even bigger number. If I evaluate this, Let's say just 10, It gives me just four. Yeah. So as the input gets bigger, the output is getting bigger. That kind of makes sense because I've got a number that's growing in the numerator, divided by a number that's growing but not as quickly in the denominator log grows more slowly than the number itself. So I've got a large number divided by a less large number, which means that this is also going to approach positive infinity but gradually. So as we head off in this direction, we're approaching infinity gradually. So somewhere in the middle here there is a point where it decreases and then increases again. And that is sort of a rough sketch of what this will look like. Let's just see how close we got. Hope I made it come down too far. Oops. So this side is approaching infinity. This side is approaching infinity gradually I got the slopes wrong and I got the minimum value wrong but must still relatively accurate given the information that we have. Mhm

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James Stewart

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Heather Zimmers

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals