Evaluating the Limit of $p$^1/n

The problem asks for the value of the limit: $ L = \lim_{n \to \infty} \sqrt[n]{p} $ given the condition that $p > 0$. We can rewrite the $n$-th root using exponents:

$ \sqrt[n]{p} = p^{1/n} $ So the limit becomes:

$ L = \lim_{n \to \infty} p^{1/n} $

Step-by-Step Limit Derivation

To determine the value of this limit, consider the behavior as $n$ becomes very large. The exponent $1/n$ approaches $0$. Since $p$ is a positive constant, we are looking at $p$ raised to a power approaching $0$.

Alternatively, we can use logarithms for a formal derivation. Let $y = p^{1/n}$. Take the natural logarithm of both sides:

$ \ln(y) = \ln(p^{1/n}) $

Using the power rule for logarithms ($\ln(a^b) = b \ln(a)$):

$ \ln(y) = \frac{1}{n} \ln(p) $

Now, find the limit of $\ln(y)$ as $n \to \infty$:

$ \lim_{n \to \infty} \ln(y) = \lim_{n \to \infty} \left( \frac{1}{n} \ln(p) \right) $

Because $p$ is a constant, $\ln(p)$ is also a constant. We know that $\lim_{n \to \infty} \frac{1}{n} = 0$. Thus:

$ \lim_{n \to \infty} \ln(y) = (\ln(p)) \times \left( \lim_{n \to \infty} \frac{1}{n} \right) = \ln(p) \times 0 = 0 $

Since the limit of $\ln(y)$ is 0, the limit of $y$ is $e$ raised to the power of this limit:

$ L = \lim_{n \to \infty} y = e^{\lim_{n \to \infty} \ln(y)} = e^0 $

Any positive number raised to the power of 0 equals 1:

$ L = 1 $

Final Result

Therefore, for any $p > 0$, the limit $\lim_{n \to \infty} \sqrt[n]{p}$ is equal to $1$.