anonymous
  • anonymous
L'Hopital's lim x->-inf (1-(3/x))^(2x) Im stuck at lim """ 2xln((x-3)/x))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
change 2x to (2x)^(-1) maybe to make this a fraction
roadjester
  • roadjester
Is this \[\lim_{x \rightarrow -\infty}{\frac 1{3x}}^{2x}\]?
amistre64
  • amistre64
\[\lim \frac{ln(1-\frac{3}{x})}{1/2x}\]

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anonymous
  • anonymous
how do you explain that mathematically? divide by 2x^2?
amistre64
  • amistre64
\[\frac{1}{a}=a^{-1}\]
anonymous
  • anonymous
that gives you 0/(1/0) does lhop work for that?
anonymous
  • anonymous
1/inf*
amistre64
  • amistre64
its indeterminant :) but id have to chk to be sure
roadjester
  • roadjester
0/(1/0) is 0/infinity
roadjester
  • roadjester
L'hopital's only works for 0/0 or infinity/infinity
anonymous
  • anonymous
yeah so how do I know to stop there? I mean couldn't we manipulate it to get 0/0 or inf/inf?
roadjester
  • roadjester
Yes, what Amistre64 wrote does get you that
roadjester
  • roadjester
3/infinity is 0 ln(1)=0
amistre64
  • amistre64
1-3/inf = 1; ln(1) = 0
anonymous
  • anonymous
I get that part but can't I do something else to make it so its 0/0 or something
roadjester
  • roadjester
It is already 0/0
anonymous
  • anonymous
with what amistre wrote we have 0/ (1/-inf)) right?
amistre64
  • amistre64
i spose you can do anything you want to it; but why would some other thing have to be worked out if we got something that works now?
roadjester
  • roadjester
yes
amistre64
  • amistre64
1/2x = 1/inf = 0
anonymous
  • anonymous
I was under the impression that we determined that we can't use l'hops rule.
roadjester
  • roadjester
we can
roadjester
  • roadjester
Let me write it out shall I?
anonymous
  • anonymous
so it is 0/0 ahh, my apologies calc teachers here don't teach us anything very useful like we can use 1/inf lol
amistre64
  • amistre64
lim ln(1-3/-inf) -0 x>-inf --------- = ---- if anything :) 1/2(-inf) -0
anonymous
  • anonymous
this is a huge mess..
roadjester
  • roadjester
|dw:1333382582070:dw|
anonymous
  • anonymous
this is so frustrating they make these derivatives too much work
roadjester
  • roadjester
Not really. try integrating ln(x^2-x+2)
anonymous
  • anonymous
We aren't that far, I can't integrate on this test. God I hate calculus. I loved precalc. This semester ruined math for me. what do we do now that we have (3/x^2)/(1-3/x)/(1/4x^2). thats 0/1/0 0/1= 0 so we can do it again?
roadjester
  • roadjester
You flip the fraction.
roadjester
  • roadjester
You flip the fraction.\[\frac{{\frac 3 {x^2}}\over {1-\frac 3 x}} {\frac1 {4x^2}}\]
roadjester
  • roadjester
\[{{\frac 3 {x^2}}\over {1- \frac 3 x}}\times\frac {{4x}^2} 1\]
roadjester
  • roadjester
and remember to take your answer to the power of e. You used natural log so you need to do the opposite to get y back. otherwise your answer is the limit of lny
anonymous
  • anonymous
e^12?
anonymous
  • anonymous
hmm thats not right
roadjester
  • roadjester
well, do you have an answer in the back of your book or something?
anonymous
  • anonymous
says wolfram
roadjester
  • roadjester
According to my calculator, it should be e^(-6)
anonymous
  • anonymous
yeah thats right but how do we get that from what we just got?
roadjester
  • roadjester
we don't. We may have made a mistake somewhere.
anonymous
  • anonymous
awwh the bottom derive is -1/2x^2 found it!
anonymous
  • anonymous
I think you just saved my calc grade roadjester lol I totally get how to l'hops rule with this ish now!
roadjester
  • roadjester
sure, no problem. And sorry about the wrong denominator derivative. I was helping someone else with synthetic division and my connection keeps getting cut. Jumping back and forth isn't that easy this morning for some reason...oh well.
anonymous
  • anonymous
its cool! thanks so much for your help
anonymous
  • anonymous
I've had an off week I just got done studying for 40 hours for a physics exam and switch back to calc is a huge pain >>

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