<p>Hey. I was wondering if you could help with this BC problem. We've been working on L'Hopital's rule, and I can't seem to figure out what do do for this...</p>
<p>lim (e^(5/x)-3x)^(x/2)
x->0</p>
<p>Thanks for your help!</p>
<p>Are you sure you have to use L'Hopital's Rule for this one?</p>
<p>If you just multiply the powers then you'll get:</p>
<p>lim e^((5/2)-(3/2)x^2)
x->0</p>
<p>which would be e^(5/2).</p>
<p>I haven't used L'Hopital's Rule in awhile (I'm in AB and our teacher just showed it to us as an easier way to solve problems involving limits), so otherwise I don't know how you would do that.</p>
<p>find limit of the natural log. then raise e to the limit you found?</p>
<p>it could help to split the exponents out first... (e^2.5-(3x^2)/2) then... you take the natural log of both sides so as to single out the e, u'd have ln(y)=(2.5-(3x^2))/2 which would have a limit on the right which would be 2.5/2 then e^1.25? maybe??? that's what cherrybarry basically said... but i don't think that's right, it's supposed to be in an indeterminate form to use l'hopital's rule, and i don't see how u can manipulate that to 0/0 or infin/infin...</p>