<p>K so this is a math question straight out of College Board's Math 1and2 Subject Tests Book.</p>
<p>What value does (ln x)/(x-1) approach as x approaches 1?
A. 0
B. 0.43
C. 1
D. 2
E. It does not approach a unique value.</p>
<p>This is question 31 from the first test in Math level2 in the book. The explanation says it is C. It also goes on about how if you use a graphing calculator you can see the function approaches 1 as x approaches 1 from both sides...it says you can examine a table of values...</p>
<p>But i answered E, because if x approaches 1, then the function approaches division by zero, which is an undefined value. i even graphed it on my graphing calculator (TI-84 Plus Silver) and took a look on the table of values. It says when x is 1 y is 'error'. That means that the number the function approaches, when x approaches 1, is E (not a unique value)...</p>
<p>k yea but i do not see how that is possible, eh? like division by zero, that is undefined. and plus my graphing calculator said so too…any1 got like an explanation?</p>
<p>The actual point (1,1) isn’t on the function, but every point around it is there. It uses the word “approaches” to get around this.
1.00000001 and .99999999 are both defined.</p>
<p>Yea 1 is a unique value, but the value 1 is not part of the function. It is not defined, so where 1 should be is a value that is not unique; an undefined value.</p>
<p>edit: you might also notice that if x=1, then lnx equals 0. maybe collegeboard is getting the answer 1 from 0/0, because anything divided by itself is 1, right? but then i also heard of other rules too, like 0 divided by anything is 0. also, anything divided by 0 is undefined…so is 0/0 equal to 1, 0, or undefined?</p>
<p>Use l’hopital’s rule. When both the top and bottom both approach 0 or both approach infinity, take the derivative of the numerator and denominator. Hence, (1/x)/1 == 1/x. Then, plug in x == 1. By this method, you get limit as x approaches 1 to be 1.</p>
<p>Taking the limits of functions is actually part of intro to calculus stuff (like, that the limit of the rate of change of y divided by the rate of change of x^2 as the change in x approaches 0 equals 2X is the same thing as saying y = x^2 therefore dy/dx = 2x. If you’re not in calc, it’s okay if you barely got that. Plus, it’s hard to describe math in regular language) </p>
<p>Whenever you see that phrasing (blah blah “approach” blah) you need to remember that the important thing isn’t that the function equates to 0/0 or 0^0 with the given X value. Instead, do like lockn proposed and look at the values directly to the left and right of it.</p>
<p>…whaa? i do not get this. where is everyone getting this (1/x)/(1)? the question in the book was (ln x)/(x-1)…could someone take me through this step by step?</p>
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<p>i am not in intro calc, so i really do not get this stuff. So what does MyPseudonym mean by ‘look at the values directly to the left and right of it’? I searched up this l’hospital’s rule and i do not get it.</p>
<p>I really need someone to go over this step by step. Thanks a lot.</p>
<p>but then why can the answer not be E? because 1 is not in the function, and because at the value 1 the function is undefined, you can also say that the value the function is approaching is not unique…rite?</p>
<p>You are having trouble because you are fixating on what happens to the function when you get all the way to the value x=1. But limits don’t work that way. When a limit question asks: “What value does f(x) approach as x approaches some number?” we do not care at all about what happens when you get all the way there! As earlier posts have said, all that matters is this: is there a value that f(x) seems to get CLOSE to as x gets CLOSE to that input number.</p>
<p>In this case, x approaches 1. So check out 1.0001 and .999 – do they give an answer in the same neighborhood? Yes – both f(1.0001) and f(.999) seem to be close to 1.</p>
<p>That f(1) itself is undefined has no effect on the value of the limit.</p>