<p>Hey guys can somebody help me with these two math 2c problems? I'm taking it tomorrow!</p>
<p>What value does ln x/(x-1) approach as x approaches 1?</p>
<p>If a and b are in the domain of a function f and f(a) is less than f(b), which of the following must be true?
A. a=0 or b=0
B. a is less than b
C. a is greater thanb
D. a does not equal b
E. a equals b</p>
<p>If n distincet planes intersect in a line, and another line, l, intersects one of these planes in a single point, what is the least number of these n planes that l could intersect?</p>
<p>For the second one, is it possible to prove that a is less than b?</p>
<p>I don’t get why the least number of these n planes that l could intersect is n-1 though because isn’t the least number n, where there would be no parallel lines?</p>
<p>It is impossible to prove anything about a and b except that they are not equal. Suppose you have all f(x)'s where a<b…now all=“” -f(x)'s=“” have=“” a=“”>b. f(x) is not specified.</b…now></p>
<p>l intersects one of the planes in one point, so it does not lie in the plane. Imagine the planes, as well as the line that they all intersect in, as if they were a fan you were looking at from below. like this, where the line is the dot and there are 4 planes. Supposing l went<br>
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through one of the planes and l is parallel to one of the other planes. Because all the planes are distinct, no two planes are parallel, so l cannot intersect less than n-1.</p>
<p>i’m sorry to keep asking the same question but…
now…i got your point but i don’t get why it can’t be n-2 and not n-1…</p>
<p>and is it possible to graph ln(1)/0 on the TI-89 because the answer is supposed to be 1, not -1.</p>
<p>for the f(a) and f(b) problem, i don’t really know what it means by f(a) is less than f(b)…Is it just that the value of y when any x is plugged in, is a greater value?</p>
<p>Sorry to be so annoying but i really appreciate your help! :)</p>
<p>Find lim x-> 1 of ln x/(x-1). First off, I’m assuming you mean lim x-> 1 of [ln x]/(x-1) and not lim x-> 1 of ln [x/(x-1)]. .</p>
<p>There’s an easy way to do this with calculus. However, you don’t need it. What you can do is take the limit from both sides. If they are not equal the limit DNE. So: </p>
<p>lim x-> 1- of ln x/(x-1) = 1</p>
<p>and</p>
<p>lim x-> 1+ of ln x/(x-1) = 1 </p>
<p>To evaluate the above two (you’re probably asking: how did you get those equaled 1?) plug in a value slightly above and slightly below 1 (like .9999 and 1.0001). This isn’t entirely mathematically sound (for example, the function could have strange behavior and do crazy stuff between 1 and 1.0001 that you wouldn’t catch with this). But just by eyeballing the function, it’s not going to happen.</p>