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Problem 9 Medium Difficulty

(a) Use a graph to find a number $ \delta $ such that
if $ 2 < x < 2 + \delta $ then $ \dfrac{1}{\ln (x - 1)} > 100 $

(b) What limit does part (a) suggest is true?

Answer

(a) $\delta \approx .01$
(b) From part (a), we see that as $x$ gets closer to 2 from the right, $y$ increases without bound. In symbols,
$\lim _{x \rightarrow 2^{+}} \frac{1}{\ln (x-1)}=\infty$

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Video Transcript

This is problem number nine of the Stuart Calculus eighth edition Section 2.4 party. Use a grass to find a number of Delta such that if two is less than X, is less than two plus delta, then one over the natural log of the quantity exponents one is greater than 100. Okay, so we're going to answer this directly. Basically, we're looking for this region between two non, including two and two, plus a very small number of delta. So X can be any number between here. And we need to set this limit such that the values for X all proceeded to give a value greater than 100 on this graph. So we will take this function for now. This is our function one divided by log rhythm of X minus one. And if we throw that into a graphing calculator, we should see something here similar to at two. There's an ask them to the function looks like this. And then, at this point, let's say it's 100 and we're looking for this limiting X value, which is value two plus delta away. First, it's close to two, but it's not. Add too. Okay, So what we're going to do is we are going to essentially, uh, solve this problem here for X. So we're going to take this and multiply both sides by the denominator here. So we're going to get one greater than 100 times. Ellen, The quantity express one divided by 100. That gives this 1000.1 is creator than Alan Times this quantity mhm here in in calculator. The next step to isolate the X will be to take the exponent on E of both sides of 0.1 greater than e to the Ln of X minus one. These are inverse operations, so they will cancel. The value to the left is approximately one point 01 greater than X minus one. Independently, we add X or plus one to both sites. And here we see that X is less than 2.1. And we have determined that this delta is a point to one. Since this value here that gives us 100 is 0.1 greater than two. So basically, this is two plus zero points are one. And this is what our delta is equal to Approximately 0.1 So anything less than this value. So anything less than 2.1 gives you. As you can tell, the function goes much greater than 100. So it satisfies this case, and we've found that limiting case any value greater than 0.1 away from two. So 2.2 gives us values of this function less than 100. And that's not what we're looking for. So in the end, what this tells us is that for party, what limit does this part a suggest is true? We will rewrite, Um, we will rewrite the limit. We're going to say that the limit as X approaches to of this natural log of X minus one is equal to zero. And we're going to show why this is true. Because this first inequality, when we subtract two to both sides, gives us zero or two to each term gives a zero less. The next one is to less than Delta or the absolute value of X men is to is less than this delta value here. As we rearranged it, we saw that we were really dealing with the function Ln of X minus one. Some of the other part of the definition is that the absolute value and a Ellen of the quantity express one minus zero, so it's distance from zero is less than some absolute value. So we see that because this is the way that it is. The limit suggested from party is that the limited expertise to have this function? Allen of expand as one approaches zero or is equal to zero, and that's where we get in this case.