Question

The value of the following limit is \(\underline{\hspace{1cm}}\). \[ \lim_{x \to 0^+} \frac{\sqrt{x}}{1 - e^{2\sqrt{x}}} \]

Hint:

For small \( x \), use approximations such as \( e^{2\sqrt{x}} \approx 1 + 2\sqrt{x} \) to simplify limits involving exponential functions.

Answer 1

Correct Answer: -0.5

To evaluate the limit, we use the standard series expansion of the exponential function near zero: \[ e^{y} = 1 + y + \frac{y^2}{2!} + \cdots \] Here, let \( y = 2\sqrt{x} \). As \( x \to 0^+ \), we have \( y \to 0 \).

Using the first-order approximation: \[ e^{2\sqrt{x}} \approx 1 + 2\sqrt{x} \]

Substitute this into the given expression: \[ \frac{\sqrt{x}}{1 - e^{2\sqrt{x}}} \approx \frac{\sqrt{x}}{1 - (1 + 2\sqrt{x})} = \frac{\sqrt{x}}{-2\sqrt{x}} \]

Simplifying: \[ = -\frac{1}{2} \]

Thus, the value of the limit is: \[ \boxed{-0.5} \]