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!
To evaluate the limit lim(x->0+) (e^(2x) + x)^(1/3x), where you cannot directly apply L'Hopital's rule or logarithmic differentiation, you can try rewriting the expression in a different form.
Let's begin by taking the natural logarithm of both sides of the expression. This allows us to bring the exponent down and convert the power into a multiplication:
ln[lim(x->0+) (e^(2x) + x)^(1/3x)] = lim(x->0+) ln[(e^(2x) + x)^(1/3x)]
Now, we can use the property of logarithms that states ln(a^b) = b ln(a):
lim(x->0+) [1/3x * ln(e^(2x) + x)] [Applying the property of logarithms]
= 1/3 * lim(x->0+) [ln(e^(2x) + x) / x ]
Now, we can use a known limit identity lim(x->0) ln(1+x) / x = 1:
lim(x->0+) [1/3 * (ln(e^(2x) + x) / x)] [Splitting the limit]
= 1/3 * lim(x->0+) [ln(e^(2x) + x) / x] [Using the known limit identity]
= 1/3 * 1 [Applying the limit identity]
So, the final result is:
lim(x->0+) (e^(2x) + x)^(1/3x) = 1/3
Therefore, the limit of (e^(2x) + x)^(1/3x) as x approaches 0 from the right is 1/3.