L’Hospital’s Rule and Indeterminate Forms — Question 4

Question 4 - Solution

We identify the limit as the indeterminate form: $$\lim_{x \to 0^+} \left(1 + 2x\right)^{\frac{1}{x}} = 1^\infty$$

Let: $$y = \left(1 + 2x\right)^{\frac{1}{x}} \Rightarrow \ln y = \frac{1}{x} \ln(1 + 2x)$$

Now compute the limit: $$\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \frac{\ln(1 + 2x)}{x}$$

This is a $\frac{0}{0}$ form. Apply L’Hospital’s Rule:

$$= \lim_{x \to 0^+} \frac{\frac{d}{dx} \ln(1 + 2x)}{\frac{d}{dx} x} = \lim_{x \to 0^+} \frac{\frac{2}{1 + 2x}}{1} = \frac{2}{1} = 2$$

So: $$\ln y \to 2 \quad \Rightarrow \quad y \to e^2$$

Final Answer: $$\boxed{e^2}$$