Problems Based on Limits of the Form $x \to \infty$

When we talk about limits of the form $x \to \infty$, we are interested in understanding the behavior of a function as the independent variable $x$ grows without bound. This concept is crucial in calculus and mathematical analysis, as it helps us to understand the asymptotic behavior of functions.

Understanding Limits at Infinity

The notation $\lim_{x \to \infty} f(x) = L$ means that as $x$ becomes arbitrarily large, the function $f(x)$ approaches the value $L$. If $L$ is a finite number, we say that $f(x)$ has a horizontal asymptote at $y = L$. If $L$ is $\infty$ or $-\infty$, the function increases or decreases without bound.

Common Forms and Results

Here are some common forms and results for limits at infinity:

Form	Result	Condition
$\lim_{x \to \infty} \frac{1}{x^n}$	$0$	$n > 0$
$\lim_{x \to \infty} \frac{a}{x}$	$0$	$a$ is a constant
$\lim_{x \to \infty} \frac{ax^n}{bx^m}$	$\frac{a}{b}$	$n = m$
$\lim_{x \to \infty} \frac{ax^n}{bx^m}$	$\infty$ or $-\infty$	$n > m$
$\lim_{x \to \infty} \frac{ax^n}{bx^m}$	$0$	$n < m$
$\lim_{x \to \infty} (1 + \frac{1}{x})^x$	$e$	-

Techniques for Evaluating Limits at Infinity

When evaluating limits of the form $x \to \infty$, we often use the following techniques:

1. Factoring and Simplifying: Factor out the highest power of $x$ in the numerator and denominator, then simplify.
2. L'Hôpital's Rule: If the limit is of the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$, differentiate the numerator and denominator separately and then take the limit.
3. Substitution: For complex expressions, sometimes a substitution can simplify the limit.
4. Asymptotic Comparison: Compare the function to a simpler function with known behavior at infinity.

Examples

Let's go through some examples to illustrate these points.

Example 1: Polynomial over Polynomial

Evaluate $\lim_{x \to \infty} \frac{3x^2 + 2x + 1}{4x^2 - 5x + 2}$.

Solution:

Factor out the highest power of $x$ in the numerator and denominator:

$$ \lim_{x \to \infty} \frac{3x^2 + 2x + 1}{4x^2 - 5x + 2} = \lim_{x \to \infty} \frac{x^2(3 + \frac{2}{x} + \frac{1}{x^2})}{x^2(4 - \frac{5}{x} + \frac{2}{x^2})} = \lim_{x \to \infty} \frac{3 + \frac{2}{x} + \frac{1}{x^2}}{4 - \frac{5}{x} + \frac{2}{x^2}}. $$

As $x \to \infty$, the terms with $x$ in the denominator approach $0$, so we have:

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

Example 2: Exponential Function

Evaluate $\lim_{x \to \infty} e^{-x}$.

Solution:

As $x \to \infty$, $-x \to -\infty$, and we know that $e$ raised to any negative power that goes to infinity approaches $0$. Therefore:

$$ \lim_{x \to \infty} e^{-x} = 0. $$

Example 3: Using L'Hôpital's Rule

Evaluate $\lim_{x \to \infty} \frac{\ln(x)}{x}$.

Solution:

This is of the form $\frac{\infty}{\infty}$, so we can apply L'Hôpital's Rule:

$$ \lim_{x \to \infty} \frac{\ln(x)}{x} = \lim_{x \to \infty} \frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x)} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0. $$

Example 4: Asymptotic Comparison

Evaluate $\lim_{x \to \infty} \frac{\sin(x)}{x}$.

Solution:

We know that $-1 \leq \sin(x) \leq 1$ for all $x$. Therefore, $-\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$. As $x \to \infty$, both $-\frac{1}{x}$ and $\frac{1}{x}$ approach $0$. By the Squeeze Theorem, we have:

$$ \lim_{x \to \infty} \frac{\sin(x)}{x} = 0. $$

Conclusion

Understanding limits as $x \to \infty$ is essential for analyzing the long-term behavior of functions. By using techniques such as factoring, L'Hôpital's Rule, substitution, and asymptotic comparison, we can evaluate these limits and gain insight into the properties of functions.