The problem is to evaluate the limit:

$\lim_{n \to \infty} \left(\frac{4n+1}{4n+3}\right)^{2n-1}.$

Step 1: Simplify the base of the exponential

The base is $\frac{4n+1}{4n+3}$. Divide numerator and denominator by $n$ to simplify: $\frac{4n+1}{4n+3} = \frac{4 + \frac{1}{n}}{4 + \frac{3}{n}}.$

As $n \to \infty$, $\frac{1}{n} \to 0$ and $\frac{3}{n} \to 0$. Therefore, the base approaches: $\frac{4 + 0}{4 + 0} = 1.$

Step 2: Approximation for small deviations from 1

Let: $x_n = \frac{4n+1}{4n+3}.$ We rewrite $x_n$ as: $x_n = 1 - \frac{2}{4n+3}.$

This approximation holds because the difference from 1 is small as $n$ becomes large.

Step 3: Approximation of the exponential

We now focus on: $x_n^{2n-1} = \left(1 - \frac{2}{4n+3}\right)^{2n-1}.$

Using the logarithmic expansion for small deviations: $\ln(1 - u) \approx -u \quad \text{for small } u.$

Here, $u = \frac{2}{4n+3}$, so: $\ln(x_n) \approx -\frac{2}{4n+3}.$

Thus: $\ln(x_n^{2n-1}) = (2n-1) \cdot \ln(x_n) \approx (2n-1) \cdot \left(-\frac{2}{4n+3}\right).$

Simplify the expression: $\ln(x_n^{2n-1}) \approx -\frac{2(2n-1)}{4n+3}.$

Step 4: Evaluate the limit

As $n \to \infty$, approximate: $\frac{2(2n-1)}{4n+3} \approx \frac{4n}{4n} = 1.$

Thus: $\ln(x_n^{2n-1}) \to -1.$

Exponentiating both sides: $x_n^{2n-1} \to e^{-1}.$

Final Answer:

$\lim_{n \to \infty} \left(\frac{4n+1}{4n+3}\right)^{2n-1} = \frac{1}{e}.$

Would you like a deeper breakdown of any step? Let me know!

Related Questions:

1. How can logarithmic approximations be applied to other limit problems?
2. Why does the approximation $\ln(1 - u) \approx -u$ work for small $u$?
3. What happens if the exponent grows faster (e.g., $3n^2$) in similar limits?
4. How does this problem relate to the exponential function's definition?
5. What is the geometric interpretation of $e^{-1}$ in this context?

Tip:

When solving exponential limits, approximating small changes using logarithms is a powerful technique!