Question

If \(a\gt 0,\ b\gt 0\), then \[ \lim_{n\to\infty} \left( \frac{a+b^{1/n}-1}{a} \right)^n = \] is equal to:

• A. \(a^b\)
• B. \(b^a\)
• C. \(b^{1/a}\)
• D. \(a^{1/b}\)

Hint:

Use the standard limit \[ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x \] and the expansion \[ b^{1/n}=e^{\frac{\log b}{n}}. \]

Answer 1

The Correct Option is C

Step 1: Recognize the $ 1^\infty $ form.
As $ n\to\infty $, $ b^{1/n}\to 1 $ so the base $ \to \frac{a}{a}=1 $ while the exponent $ \to \infty $. This is the $ 1^\infty $ indeterminate form.
Step 2: Rewrite the base as $ 1 + \text{small} $.
\[ L = \lim_{n\to\infty}\!\left(1+\frac{b^{1/n}-1}{a}\right)^n \] Let $ u_n = \frac{b^{1/n}-1}{a} \to 0 $.
Step 3: Apply the standard limit $ \lim(1+u)^{1/u}=e $.
\[ L = \lim_{n\to\infty}\left[(1+u_n)^{1/u_n}\right]^{n\cdot u_n} \to e^{\lim_{n\to\infty} n\cdot u_n} \]
Step 4: Compute $ \lim_{n\to\infty} n(b^{1/n}-1) $.
Write $ b^{1/n}=e^{(\ln b)/n} $. Using $ e^t-1\approx t $: $ b^{1/n}-1\approx\frac{\ln b}{n} $. So $ n(b^{1/n}-1)\to\ln b $.
Step 5: Compute the exponent limit.
\[ \lim_{n\to\infty} n\cdot u_n = \frac{\ln b}{a} \]
Step 6: Exponentiate to find L.
\[ L = e^{(\ln b)/a} = b^{1/a} \] \[ \boxed{b^{1/a}} \]