Differential Calc (Math)
posted by Joey on .
Stuck... can't seem to use l'hopitals, nor can I use logrithmic differentiation...
lim (e^(2x)+ x)^(1/3x)
x>0+
Thank you so much

If that's ^(1/3)x, then you have no problem:
(e^0+0)^(0) = (1+0)^0 = 1^0 = 1
Now, if it's ^(1/(3x)) then you have
1^∞, so we have to get creative.
How about:
e^2x = 1 + 1/1(2x) + 1/2(2x)^2 + ...
= 1+2x as x>0 because the higher powers become insignificant.
lim (1+3x)^1/(3x) = lim (1+u)^1/u = e
Can't think of something more rigorous at the moment. 
I'm not quiet sure how e^2x equals what you put,
and
lim (1+3x)^1/(3x) = lim (1+u)^1/u = e
doesn't seem to make sense either..
Maybe a more dumbed down version? And it is in fact ^1/(3x) 
the Taylor series for e^u = 1 + 1/1! u + 1/2! u^2 + 1/3! u^3 + ...
the definition of e is lim(u>0) (1+u)^1/u 
Hm... Never learned this before, so I'm not quite sure if I can use this.
However, I appreciate the effort! 
How about this:
lim(x>0) u^v = lim(x>0) e^(v ln u)
here we have
u = e^2x + x
v = 1/(3x)
lim e^(ln(e^2x+x)/3x)
= e^ [lim(ln(e^2x+x)/3x)]
now use l'Hospital's Rule to see that
lim ln(e^2x+x)/3x = (2e^2x+1)/(e^2x+x) / 3
= (2+1)/(0+1) / 3
= 3/3 = 1
and our limit is now
e^(1) = e 
WOW! Thank you so much! You rock!