Limits At Infinity, Part II — Question 8

Question 8 - Solution

We evaluate $$\lim_{x \to \infty} x^2\left(\sqrt{x^2 + 4x} - x\right).$$

As $x \to \infty$, both $\sqrt{x^2 + 4x}$ and $x$ approach infinity, so the expression inside the parentheses has the indeterminate form $\infty - \infty$. Multiplying by $x^2$ significantly changes the behavior of the limit, so careful analysis is required.

Step 1: Rationalize the Expression

We first rationalize the difference: $$\sqrt{x^2 + 4x} - x = \frac{(\sqrt{x^2 + 4x} - x)(\sqrt{x^2 + 4x} + x)}{\sqrt{x^2 + 4x} + x}.$$

Using the identity $(a-b)(a+b)=a^2-b^2$, this becomes $$= \frac{(x^2 + 4x) - x^2}{\sqrt{x^2 + 4x} + x} = \frac{4x}{\sqrt{x^2 + 4x} + x}.$$

Substitute into the original expression: $$\lim_{x \to \infty} x^2 \cdot \frac{4x}{\sqrt{x^2 + 4x} + x} = \lim_{x \to \infty} \frac{4x^3}{\sqrt{x^2 + 4x} + x}.$$

Step 2: Simplify

Factor $x$ out of the square root: $$\sqrt{x^2 + 4x} = x\sqrt{1 + \frac{4}{x}} \quad (x>0).$$

Then $$\frac{4x^3}{x\left(\sqrt{1 + \frac{4}{x}} + 1\right)} = \frac{4x^2}{\sqrt{1 + \frac{4}{x}} + 1}.$$

Step 3: Take the Limit

As $x \to \infty$, $\frac{4}{x} \to 0$, so $$\sqrt{1 + \frac{4}{x}} \to 1.$$

Thus, $$\lim_{x \to \infty} \frac{4x^2}{\sqrt{1 + \frac{4}{x}} + 1} = \lim_{x \to \infty} \frac{4x^2}{2} = \infty.$$

Final Answer

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