Limits At Infinity, Part II — Question 10

Question 10

Evaluate the limit: $$\lim_{x \to \infty} x\left( \sqrt{x^2 + 3x} - x \right).$$

Question 10 - Solution

We evaluate $$\lim_{x \to \infty} x\left( \sqrt{x^2 + 3x} - x \right).$$

As $x \to \infty$, both $\sqrt{x^2 + 3x}$ and $x$ grow without bound, so the expression inside the parentheses has the indeterminate form $\infty - \infty$. Multiplying by $x$ changes the rate at which the expression grows, so careful analysis is required.

Step 1: Rationalize the Expression

We first rationalize the difference: $$\sqrt{x^2 + 3x} - x = \frac{(\sqrt{x^2 + 3x} - x)(\sqrt{x^2 + 3x} + x)}{\sqrt{x^2 + 3x} + x}.$$

Using the identity $(a-b)(a+b) = a^2 - b^2$, this becomes $$= \frac{(x^2 + 3x) - x^2}{\sqrt{x^2 + 3x} + x} = \frac{3x}{\sqrt{x^2 + 3x} + x}.$$

Substitute this into the original limit: $$\lim_{x \to \infty} x \cdot \frac{3x}{\sqrt{x^2 + 3x} + x} = \lim_{x \to \infty} \frac{3x^2}{\sqrt{x^2 + 3x} + x}.$$

Step 2: Simplify

Factor $x$ out of the square root: $$\sqrt{x^2 + 3x} = x\sqrt{1 + \frac{3}{x}} \quad (x > 0).$$

Then $$\frac{3x^2}{x\left(\sqrt{1 + \frac{3}{x}} + 1\right)} = \frac{3x}{\sqrt{1 + \frac{3}{x}} + 1}.$$

Step 3: Take the Limit

As $x \to \infty$, $\frac{3}{x} \to 0$, so $$\sqrt{1 + \frac{3}{x}} \to 1.$$

Therefore, $$\lim_{x \to \infty} \frac{3x}{\sqrt{1 + \frac{3}{x}} + 1} = \lim_{x \to \infty} \frac{3x}{2} = \infty.$$

Final Answer

$$\boxed{\text{The limit diverges to } \infty.}$$