Limits At Infinity, Part II — Question 5

Question 5 - Solution

We are given: $$\lim_{x \to \infty} \left( \frac{x^2 + 3x - 7}{\sqrt{x^4 + 2x^3 + 5}} \right)$$

Step 1: Understand the growth of numerator and denominator

As $x \to \infty$, both numerator and denominator tend to infinity, so we need to simplify.

Step 2: Factor out highest power of $x$

In the numerator: highest power is $x^2$ In the denominator (inside square root): highest power is $x^4$

Factor numerator: $$x^2 \left(1 + \frac{3}{x} - \frac{7}{x^2}\right)$$

Factor denominator inside square root: $$\sqrt{x^4 \left(1 + \frac{2}{x} + \frac{5}{x^4} \right)} = x^2 \sqrt{1 + \frac{2}{x} + \frac{5}{x^4}}$$

Step 3: Substitute factored forms into the expression**

$$\frac{x^2 \left(1 + \frac{3}{x} - \frac{7}{x^2}\right)}{x^2 \sqrt{1 + \frac{2}{x} + \frac{5}{x^4}}}$$

Cancel $x^2$:

$$\frac{1 + \frac{3}{x} - \frac{7}{x^2}}{\sqrt{1 + \frac{2}{x} + \frac{5}{x^4}}}$$

Step 4: Take the limit as $x \to \infty$

As $x \to \infty$, all terms with $\frac{1}{x}$ go to 0:

$$\lim_{x \to \infty} \frac{1 + \frac{3}{x} - \frac{7}{x^2}}{\sqrt{1 + \frac{2}{x} + \frac{5}{x^4}}} = \frac{1 + 0 - 0}{\sqrt{1 + 0 + 0}} = \frac{1}{1} = 1$$

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