L’Hospital’s Rule and Indeterminate Forms — Question 3

Question 3 - Solution

We observe that as $x \to \infty$:

- $\ln\left(1 + \frac{1}{x} \right) \to 0$ - $x \to \infty$

So the expression becomes: $$\infty \cdot 0 \quad \text{(indeterminate form)}$$

We rewrite it as a quotient: $$x \ln\left(1 + \frac{1}{x} \right) = \frac{\ln\left(1 + \frac{1}{x} \right)}{1/x}$$

Now as $x \to \infty$, we get a $\frac{0}{0}$ form, so we can apply L’Hospital’s Rule.

Let’s compute the limit using L’Hospital:

$$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x} \right)}{1/x} = \lim_{x \to \infty} \frac{\frac{d}{dx} \ln\left(1 + \frac{1}{x} \right)}{\frac{d}{dx} \left( \frac{1}{x} \right)}$$

Differentiate numerator and denominator:

- Numerator derivative: $$\frac{d}{dx} \ln\left(1 + \frac{1}{x} \right) = \frac{-1}{x^2 (1 + \frac{1}{x})} = \frac{-1}{x(x+1)}$$

- Denominator derivative: $$\frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2}$$

Now plug in:

$$= \lim_{x \to \infty} \frac{-1/(x(x+1))}{-1/x^2} = \lim_{x \to \infty} \frac{x^2}{x(x+1)} = \lim_{x \to \infty} \frac{x}{x+1}$$

$$\lim_{x \to \infty} \frac{x}{x+1} = 1$$

Final Answer: $$\boxed{1}$$