Question

If $p > 0$, then $\lim_{n \to \infty} \sqrt[n]{p}$ :

• A. 0
• B. 1
• C. $e$
• D. does not exist.

Hint:

Standard Real Analysis Limits to Remember: 1. $\lim_{n \to \infty} p^{1/n} = 1$ for any $p > 0$. 2. $\lim_{n \to \infty} n^{1/n} = 1$. 3. $\lim_{n \to \infty} (1 + \frac{x}{n})^n = e^x$.

Answer 1

The Correct Option is B

Step 1: Concept
We are evaluating the limit of the sequence $a_n = p^{1/n}$ as $n \to \infty$, where $p$ is a fixed positive real constant ($p > 0$).

Step 2: Key Formulas and Approach
Using the exponential and natural logarithm relationship: \[ p^{1/n} = e^{\ln(p^{1/n})} = e^{\frac{\ln(p)}{n}} \]

Step 3: Step-by-step Explanation

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

• Express $p^{1/n}$ in exponential form: \[ \lim_{n \to \infty} p^{1/n} = \lim_{n \to \infty} \exp\left( \frac{\ln(p)}{n} \right) \]
• Since the exponential function $f(x) = e^x$ is continuous everywhere on $\mathbb{R}$, we can pass the limit inside the function: \[ L = \exp\left( \lim_{n \to \infty} \frac{\ln(p)}{n} \right) \]
• For any fixed $p > 0$, $\ln(p)$ is a real constant. Therefore: \[ \lim_{n \to \infty} \frac{\ln(p)}{n} = 0 \]
• Substituting this back: \[ L = e^0 = 1 \]
• Hence, for any $p > 0$, $\lim_{n \to \infty} \sqrt[n]{p} = 1$.

Step 4: Final Answer
The limit of $\sqrt[n]{p}$ as $n \to \infty$ for any positive constant $p$ is 1. Thus, Option (B) is correct.