Math Snap

STEP 1

Assumptions
1. We need to find the limit as $x$ approaches $0^+$.
2. The expression given is $\frac{1}{x} - \frac{1}{e^x - 1}$.
3. We will use l'Hospital's Rule if the limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
4. If a simpler method is available, we will consider using it.

STEP 2

First, let's analyze the expression to see if it results in an indeterminate form as $x \to 0^+$.
As $x \to 0^+$:
- $\frac{1}{x} \to \infty$
- $e^x - 1 \approx x$ (using the first-order Taylor expansion of $e^x$ around 0)
- Therefore, $\frac{1}{e^x - 1} \approx \frac{1}{x}$
Thus, the expression $\frac{1}{x} - \frac{1}{e^x - 1}$ becomes $\infty - \infty$, which is an indeterminate form.

STEP 3

To resolve the indeterminate form, we will combine the terms into a single fraction:
$$\frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)}$$

STEP 4

Now, evaluate the limit of the new expression:
$$\lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)}$$ As $x \to 0^+$, both the numerator and the denominator approach 0, resulting in a $\frac{0}{0}$ indeterminate form. This allows us to apply l'Hospital's Rule.

STEP 5

Apply l'Hospital's Rule by differentiating the numerator and the denominator:
Numerator: $\frac{d}{dx}(e^x - 1 - x) = e^x - 1$
Denominator: $\frac{d}{dx}(x(e^x - 1)) = (e^x - 1) + x \cdot e^x = e^x - 1 + xe^x$

STEP 6

Apply l'Hospital's Rule:
$$\lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x}$$

Evaluate the limit:
As $x \to 0^+$, the expression simplifies to:
$$\frac{e^0 - 1}{e^0 - 1 + 0 \cdot e^0} = \frac{0}{0} + 0 = 0$$ Thus, the limit is:
$$\lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x} = 0$$ The limit is $0$.