We are given the function $f(x) = \frac{4x^2 - 5}{x^2 + 2}$ and asked to find the horizontal and vertical asymptotes using limits.

1. Horizontal Asymptotes:

To find the horizontal asymptote, we examine the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$.

$f(x) = \frac{4x^2 - 5}{x^2 + 2}$

As $x \to \infty$ or $x \to -\infty$, the highest degree terms (those involving $x^2$) dominate in both the numerator and the denominator. So, we can approximate:

$f(x) \approx \frac{4x^2}{x^2} = 4$

Thus, the horizontal asymptote is:

$y = 4$

2. Vertical Asymptotes:

Vertical asymptotes occur where the denominator is zero and the numerator is nonzero. The denominator is $x^2 + 2$.

$x^2 + 2 = 0 \quad \text{gives no real solutions because} \quad x^2 = -2 \quad \text{is not possible for real numbers}.$

Therefore, there are no vertical asymptotes.

3. Limits to Confirm Behavior:

• As $x \to \infty$, $\lim_{x \to \infty} f(x) = 4$.
• As $x \to -\infty$, $\lim_{x \to -\infty} f(x) = 4$.

These confirm the horizontal asymptote at $y = 4$.

There are no vertical asymptotes since the denominator never equals zero for real values of $x$.

Would you like more details or have any questions?

Related Questions:

1. How would the function change if there were a term like $x$ in the denominator?
2. How do slant asymptotes differ from horizontal asymptotes?
3. What if the degree of the numerator was higher than the denominator?
4. How can the limits help us determine end behavior for functions with more complex exponents?
5. Can there be horizontal asymptotes when the numerator and denominator have different degrees?

Tip:

Always compare the degrees of the numerator and denominator to quickly assess the horizontal asymptote in rational functions.