Question

The value of $\lim_{x \to \infty} (\frac{2x-1}{2x+3})^{\frac{x+1}{2}}$ is:

• A. e
• B. 0
• C. $\frac{1}{e^2}$
• D. $\frac{1}{e}$

Answer 1

Answer: (D)

$\frac{1}{e}$

Answer 2

The limit is of the form $\lim_{x \to \infty} [f(x)]^{g(x)}$, where $f(x) = \frac{2x-1}{2x+3}$ and $g(x) = \frac{x+1}{2}$. As $x \to \infty$, $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2x-1}{2x+3} = \lim_{x \to \infty} \frac{2 - 1/x}{2 + 3/x} = \frac{2}{2} = 1$. As $x \to \infty$, $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} \frac{x+1}{2} = \infty$. This is the $1^\infty$ indeterminate form.

The limit can be evaluated using the formula: $\lim_{x \to \infty} [f(x)]^{g(x)} = e^{\lim_{x \to \infty} g(x) [f(x) - 1]}$

First, calculate $f(x) - 1$: $f(x) - 1 = \frac{2x-1}{2x+3} - 1 = \frac{2x-1 - (2x+3)}{2x+3} = \frac{2x-1 - 2x - 3}{2x+3} = \frac{-4}{2x+3}$.

Next, calculate the limit of the exponent, $L = \lim_{x \to \infty} g(x) [f(x) - 1]$: $L = \lim_{x \to \infty} \left(\frac{x+1}{2}\right) \left(\frac{-4}{2x+3}\right)$ $L = \lim_{x \to \infty} \frac{-4(x+1)}{2(2x+3)}$ $L = \lim_{x \to \infty} \frac{-2(x+1)}{2x+3}$ $L = \lim_{x \to \infty} \frac{-2x-2}{2x+3}$

To evaluate this limit as $x \to \infty$, divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$: $L = \lim_{x \to \infty} \frac{\frac{-2x}{x} - \frac{2}{x}}{\frac{2x}{x} + \frac{3}{x}}$ $L = \lim_{x \to \infty} \frac{-2 - 2/x}{2 + 3/x}$ As $x \to \infty$, $2/x \to 0$ and $3/x \to 0$. $L = \frac{-2 - 0}{2 + 0} = \frac{-2}{2} = -1$.

The value of the original limit is $e^L = e^{-1}$. $e^{-1} = \frac{1}{e}$.

The final answer is $\frac{1}{e}$.