Lt (((1+×)^(1÷×))÷e)^1÷×
x->0
e=2.718281828.....
I defined e.
well, e = limit(x->0) (1+x)^(1/x), so your expressions is
(e/e)^(1/x) = 1^infinity
Hmmm. No joy there
Take a look at
http://www.wolframalpha.com/input/?i=limit+%28x-%3E0%2B%29+%28%28%281%2Bx%29^%281%2Fx%29%29%2Fe%29^%281%2Fx%29
To evaluate the limit of the expression (((1+x)^(1/x))/e)^(1/x) as x approaches 0, we can use some properties of limits and apply L'Hôpital's Rule.
L'Hôpital's Rule states that if you have an indeterminate form of 0/0 or ∞/∞, and taking the derivative of both the numerator and denominator with respect to the variable results in an expression that still produces the same indeterminate form, then the limit of the original expression is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.
Let's start by simplifying the expression:
(((1+x)^(1/x))/e)^(1/x)
Taking the natural logarithm of the expression will help us apply L'Hôpital's Rule:
ln((((1+x)^(1/x))/e)^(1/x))
Using the property of logarithms, we can bring down the exponent:
(1/x) * ln(((1+x)^(1/x))/e)
Next, we can simplify further by expanding the exponential expression inside the natural logarithm:
(1/x) * ln((1+x)^(1/x)) - (1/x) * ln(e)
Since ln(e) is equal to 1, we can simplify again:
(1/x) * ln((1+x)^(1/x)) - (1/x)
Now, let's apply L'Hôpital's Rule by taking the derivative of the numerator and denominator:
Numerator:
Using the chain rule, we have 1/x * d/dx(ln((1+x)^(1/x))).
Applying the chain rule, we get:
(1/x) * d/dx(1/x * ln(1+x)).
Simplifying further, we have:
(1/x) * (1/x * (1/(1+x))).
Combining the fractions, we get:
(1/x^2) * (1/(1+x)).
Denominator:
The derivative of 1/x is -1/x^2.
Now, we can rewrite the original expression using the derivatives:
lim(x->0) [(1/x^2) * (1/(1+x))] / [-1/x^2]
Simplifying further, we get:
lim(x->0) -1/(1+x)
Finally, plugging in x = 0, we have:
-1/(1+0) = -1/1 = -1
Therefore, the limit of the expression (((1+x)^(1/x))/e)^(1/x) as x approaches 0 is -1.