Limits At Infinity, Part II — Question 7

Question 7 - Solution

We evaluate $$\lim_{x \to \infty} x\left(\sqrt{x^2 + 5x} - x\right).$$

At first glance, as $x \to \infty$, $$\sqrt{x^2 + 5x} \to \infty \quad \text{and} \quad x \to \infty,$$ so the expression inside the parentheses has the indeterminate form $\infty - \infty$.

Step 1: Rationalize the Expression

Multiply the expression inside the parentheses by its conjugate: $$\sqrt{x^2 + 5x} - x = \frac{(\sqrt{x^2 + 5x} - x)(\sqrt{x^2 + 5x} + x)}{\sqrt{x^2 + 5x} + x}.$$

Using the identity $(a-b)(a+b)=a^2-b^2$, we obtain $$= \frac{(x^2 + 5x) - x^2}{\sqrt{x^2 + 5x} + x} = \frac{5x}{\sqrt{x^2 + 5x} + x}.$$

Thus the original limit becomes $$\lim_{x \to \infty} x \cdot \frac{5x}{\sqrt{x^2 + 5x} + x}.$$

Step 2: Simplify

Rewrite the expression as $$\lim_{x \to \infty} \frac{5x^2}{\sqrt{x^2 + 5x} + x}.$$

Factor $x$ out of the square root: $$\sqrt{x^2 + 5x} = x\sqrt{1 + \frac{5}{x}} \quad (x>0).$$

Substitute: $$\frac{5x^2}{x\left(\sqrt{1 + \frac{5}{x}} + 1\right)} = \frac{5x}{\sqrt{1 + \frac{5}{x}} + 1}.$$

Step 3: Take the Limit

As $x \to \infty$, $\frac{5}{x} \to 0$, so $$\sqrt{1 + \frac{5}{x}} \to 1.$$

Therefore, $$\lim_{x \to \infty} \frac{5x}{\sqrt{1 + \frac{5}{x}} + 1} = \lim_{x \to \infty} \frac{5x}{2} = \infty.$$

Final Answer

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