L’Hospital’s Rule and Indeterminate Forms — Question 8

Question 8 - Solution

As $x \to \infty$, we have: $$\sqrt{x^2 + x} \to \infty, \quad x \to \infty \Rightarrow \text{Indeterminate form: } \infty - \infty$$

To resolve this, multiply by the conjugate: $$\lim_{x \to \infty} \left( \sqrt{x^2 + x} - x \right) = \lim_{x \to \infty} \frac{(\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x)}{\sqrt{x^2 + x} + x}$$

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

Now simplify numerator and denominator:

Factor $x$ from the square root: $$= \lim_{x \to \infty} \frac{x}{x\left( \sqrt{1 + \frac{1}{x}} + 1 \right)} = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}$$

As $x \to \infty$, $\frac{1}{x} \to 0$, so: $$\lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$

Final Answer: $\boxed{\dfrac{1}{2}}$