<p>What value does (ln x)/(x-1) approach as x approaches 1?</p>
<p>i am going to guess it's 0.
as x gets really close to one, then the bottom will essentially be .000000000001, making any number, BUT zero, infinite)</p>
<p>but the top IS zero, since e must be raised to the zero power to be one. so it would be something like
zero divided by .0000000000001, and anything divided by or multiplied by zero is still zero</p>
<p>it's 1....................</p>
<p>What value does (ln x)/(x-1) approach as x approaches 1?</p>
<p>You can solve this problem by using L'Hopital's rule. If you plug in 1 for ln x and x-1, you get 0/0 (which is undeterminate). L'Hopital's rule says that you can find the derivative of the top and the bottom and then plug in the limit. So, the derivative of ln x is 1/x and the derivative of x-1 is simply 1. So, now we have to find the limit of x/1 as x approaches 1. Plug in 1 for x and you get 1.</p>
<p>^Haha Just learned that in school today...yeah but L'Hopital didn't really invent it...the Bernoulli brothers did and L'Hopital just basically paid/stole it...</p>
<p>l'hopitals rule</p>
<p>lim (ln x)/(x-1) = 0/0 which is indeterminate</p>
<p>Take the derivative of the top and the bottom and you get (1/x)/1. Stick 1 in for x and you get the answer 1.</p>
<p>For x->1</p>
<p>lim( (ln x)/(x-1) ) =
lim( ln x^(1/(x-1)) ) =
ln lim( x^(1/(x-1)) ) =
ln e = 1</p>
<p>wait...calc bc is on the sat? lmao</p>
<p>@gfc</p>
<p>...what...?
how does lim( x^(1/(x-1))) = e?
lol</p>
<p>Yay for L'Hopital's rule! It is 1, btw...</p>
<p>I vaguely remember you have to "e" it to get rid of the log.</p>
<p>
</p>
<p>In lay terms,
lim (approaching 1)^(approaching ∞) = e,
provided "approachings" proceed at the right paces respectively - as in
lim (1+1/n)^n = e.
n->∞</p>
<p>So
lim x^(1/(x-1)) = e
x->1</p>
<p>===================
Here's the proof.
lim (1+1/n)^n = e
n->∞</p>
<p>Substitution t=1/n yields
lim (1+t)^(1/t) = e
t->0</p>
<p>Substitution x=1+t yields
lim x^(1/(x-1)) = e</p>
<h1>x->1</h1>
<p>Or you could just pull out your old trustworthy TI-89 and feed it
limit(x^(1/(x-1)),x,1). It's still e.
lmsao</p>
<p>===================
I.dont.know, it sure looks like homework to me. ;)</p>
<p>this could be a math IIC problem although you wouldn't need do it without a calculator.</p>
<p>^precisely my point.</p>
<p>Actually, Math Level 2 does not venture into the Calc territory.
The only limits I've seen were ones of rational functions.</p>