Explanation:
The Laplace transform of a function is a powerful tool in solving differential equations. It allows us to transform a function from the time domain to the frequency domain. In this case, we are given the Laplace transform of i(t) as I(s) = 2/s(1 - s).

Laplace Transform:
The Laplace transform is defined as the integral of a function multiplied by e^(-st), where s is a complex number. In this case, the Laplace transform of i(t) is given by I(s) = ∫[i(t) * e^(-st)] dt.

Partial Fraction Decomposition:
To find the inverse Laplace transform of I(s), we need to perform a partial fraction decomposition on the expression 2/s(1 - s). This involves breaking the expression into simpler fractions that we can easily find the inverse Laplace transform of.

The partial fraction decomposition of 2/s(1 - s) is given by:
2/s(1 - s) = A/s + B/(1 - s)

To find the values of A and B, we need to solve the equation:
2 = A(1 - s) + Bs

Simplifying the equation, we get:
2 = A - As + Bs

Comparing the coefficients of s, we get:
-A + B = 0

Comparing the constant terms, we get:
A = 2

From these equations, we can solve for A and B:
A = 2
B = 2

Inverse Laplace Transform:
Now that we have the partial fraction decomposition of I(s), we can find the inverse Laplace transform by using the following table:

L^-1{1/s} = 1
L^-1{1/(1 - s)} = e^t

Using the table, the inverse Laplace transform of I(s) is given by:
i(t) = 2 + 2e^t

Limit as t ⟶∞:
As t approaches infinity, the exponential term e^t becomes very large. This means that the value of i(t) tends to be dominated by the term 2e^t. Since e^t increases exponentially, the value of i(t) also increases exponentially.

Therefore, the value of i(t) tends to be very large as t ⟶∞. The correct answer is '2', which represents the constant term in the inverse Laplace transform.

From the Final value theorem, we have