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: Simplify the expression inside the bracket.
Given limit: \[ \lim_{n\to\infty} \left( \frac{a+b^{1/n}-1}{a} \right)^n \] Rewrite the bracket: \[ \frac{a+b^{1/n}-1}{a} = 1+\frac{b^{1/n}-1}{a} \] So, the limit becomes \[ \lim_{n\to\infty} \left( 1+\frac{b^{1/n}-1}{a} \right)^n \]

Step 2: Use exponential expansion.
We know that \[ b^{1/n}=e^{\frac{\log b}{n}} \] For large \(n\), \[ e^{\frac{\log b}{n}} \approx 1+\frac{\log b}{n} \] Hence, \[ b^{1/n}-1\approx \frac{\log b}{n} \]

Step 3: Substitute in the limit.
Therefore, \[ 1+\frac{b^{1/n}-1}{a} \approx 1+\frac{\log b}{an} \] So, \[ \lim_{n\to\infty} \left( 1+\frac{\log b}{an} \right)^n \] Using \[ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x, \] we get \[ e^{\frac{\log b}{a}} \]

Step 4: Simplify the final expression.
\[ e^{\frac{\log b}{a}} = \left(e^{\log b}\right)^{1/a} \] \[ = b^{1/a} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{b^{1/a}} \]