Approach:
To find the value of the given limit, we will use L'Hôpital's Rule, which states that if the limit of the ratio of two functions is indeterminate (such as 0/0 or ∞/∞), then the limit of the ratio of their derivatives will be the same.

Calculation:
Let's start by finding the derivatives of the numerator and denominator separately:
- Derivative of (x+1) with respect to x is 1.
- Derivative of (x/e)^x√x can be found using logarithmic differentiation.
Taking the natural logarithm of (x/e)^x√x, we get:
ln((x/e)^x√x) = x√x * ln(x/e)
= x√x * (ln(x) - ln(e))
= x√x * (ln(x) - 1)
Now, taking the derivative of ln((x/e)^x√x) with respect to x, we get:
= (1/x√x) * x√x * (ln(x) - 1) + x√x * (1/x)
= ln(x) - 1 + 1
= ln(x)
Therefore, the derivative of (x/e)^x√x with respect to x is x√x * ln(x).
Now, we can rewrite the given limit as the limit of the ratio of the derivatives:
lim x→γ [1 / x√x * ln(x)]
Now, we can apply L'Hôpital's Rule by taking the derivatives of the numerator and denominator separately:
= lim x→γ [0 / (1/2√x * ln(x) + x√x * 1/x)]
= lim x→γ [0 / (1/2 * ln(x) + √x)]
= 0 / (∞)
= 0
Therefore, the value of the given limit is 0.

Conclusion:
The value of the limit lim x→γ (x+1)/(x/e)^x√x is 0.