Question

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

Answer 1

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.

Answer 2

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)

Answer 3

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

Answer 4

Hm... Never learned this before, so I'm not quite sure if I can use this.

However, I appreciate the effort!

Answer 5

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

Answer 6

WOW! Thank you so much! You rock!