Limits At Infinity, Part II — Question 1

Question 1

Evaluate the limit: $$\lim_{x \to \infty} \left( \sqrt{x^2 + 5x + 7} - \sqrt{x^2 + x} \right).$$

Question 1 - Solution

We evaluate $$\lim_{x \to \infty} \left( \sqrt{x^2 + 5x + 7} - \sqrt{x^2 + x} \right).$$

This expression has the indeterminate form $\infty - \infty$. To resolve it, we multiply by the conjugate.

Step 1: Multiply by the Conjugate

$$= \lim_{x \to \infty} \left( \sqrt{x^2 + 5x + 7} - \sqrt{x^2 + x} \right) \cdot \frac{\sqrt{x^2 + 5x + 7} + \sqrt{x^2 + x}}{\sqrt{x^2 + 5x + 7} + \sqrt{x^2 + x}}.$$

This simplifies to $$= \lim_{x \to \infty} \frac{(x^2 + 5x + 7) - (x^2 + x)}{\sqrt{x^2 + 5x + 7} + \sqrt{x^2 + x}}.$$

Step 2: Simplify the Numerator

$$(x^2 + 5x + 7) - (x^2 + x) = 4x + 7,$$ so the limit becomes $$\lim_{x \to \infty} \frac{4x + 7}{\sqrt{x^2 + 5x + 7} + \sqrt{x^2 + x}}.$$

Step 3: Factor Out $x$

Factor $x$ from each square root: $$= \lim_{x \to \infty} \frac{4x + 7} {x\left( \sqrt{1 + \frac{5}{x} + \frac{7}{x^2}} + \sqrt{1 + \frac{1}{x}} \right)}.$$

Divide numerator and denominator by $x$: $$= \lim_{x \to \infty} \frac{4 + \frac{7}{x}} {\sqrt{1 + \frac{5}{x} + \frac{7}{x^2}} + \sqrt{1 + \frac{1}{x}}}.$$

Step 4: Take the Limit

As $x \to \infty$, $$\frac{7}{x} \to 0, \quad \frac{5}{x} \to 0, \quad \frac{7}{x^2} \to 0, \quad \frac{1}{x} \to 0.$$

Thus, $$\lim_{x \to \infty} \frac{4 + \frac{7}{x}} {\sqrt{1 + \frac{5}{x} + \frac{7}{x^2}} + \sqrt{1 + \frac{1}{x}}} = \frac{4}{1 + 1} = 2.$$

Final Answer

$$\boxed{2}$$