Limits of this type are frequently encountered in the calculation of limits with polynomials or ratios of polynomials, especially when analyzing the behavior of rational functions for very large values of \(x\).
\[ \lim_{x \to \infty} \frac{1}{x^n} \]
Case \( n > 0 \)
We want to prove that:
\[ \lim_{x \to \infty} \frac{1}{x^n} = 0 \quad \text{for every } n > 0. \]
Definition of limit: for every \( \varepsilon > 0 \), there exists \( X > 0 \) such that if \( x > X \), then:
\[ \left| \frac{1}{x^n} - 0 \right| < \varepsilon. \]
But since \( \frac{1}{x^n} \) is always positive, we can rewrite:
\[ \frac{1}{x^n} < \varepsilon. \]
Now, we just need to choose \( X \) such that:
\[ X = \left(\frac{1}{\varepsilon}\right)^{\frac{1}{n}}. \]
If \( x > X \), then:
\[ x > \left(\frac{1}{\varepsilon}\right)^{\frac{1}{n}} \quad \Rightarrow \quad x^n > \frac{1}{\varepsilon} \quad \Rightarrow \quad \frac{1}{x^n} < \varepsilon. \]
Therefore, by the definition of limit, we obtain:
\[ \lim_{x \to \infty} \frac{1}{x^n} = 0. \]
Case \( n = 0 \)
If \( n = 0 \), then the expression simply becomes:
\[ \frac{1}{x^0} = 1. \]
Being constant, the limit is:
\[ \lim_{x \to \infty} \frac{1}{x^0} = 1. \]
Case \( n < 0 \)
If \( n < 0 \), we can write \( n = -m \) with \( m > 0 \). In this case:
\[ \frac{1}{x^n} = \frac{1}{x^{-m}} = x^m. \]
Since \( x^m \to \infty \) as \( x \to \infty \), it follows that:
\[ \lim_{x \to \infty} \frac{1}{x^n} = \infty. \]
Let's summarize the three cases:
- If \( n > 0 \), then \( \lim_{x \to \infty} \displaystyle \frac{1}{x^n} = 0 \).
- If \( n = 0 \), then \( \lim_{x \to \infty} \displaystyle \frac{1}{x^n} = 1 \).
- If \( n < 0 \), then \( \lim_{x \to \infty} \displaystyle \frac{1}{x^n} = \infty \).
This property is useful for studying the asymptotic behavior of rational functions.
Exercise on Notable Limit
Calculate the following limit:
\[ \lim_{x \to \infty} \frac{5x^3 + 7}{2x^5 + 4x^2 + 1}. \]
Solution. Let's divide both numerator and denominator by the highest power of \( x \) present in the denominator, which is \( x^5 \):
\[ \frac{5x^3 + 7}{2x^5 + 4x^2 + 1} = \frac{\displaystyle \frac{5x^3}{x^5} + \displaystyle \frac{7}{x^5}}{\displaystyle \frac{2x^5}{x^5} + \displaystyle \frac{4x^2}{x^5} + \displaystyle \frac{1}{x^5}}. \]
Simplifying:
\[ = \frac{5 \cdot \displaystyle \frac{1}{x^2} + 7 \cdot \displaystyle \frac{1}{x^5}}{2 + 4 \cdot \displaystyle \frac{1}{x^3} + \displaystyle \frac{1}{x^5}}. \]
Now let's apply the limit. Since we know that \( \lim_{x \to \infty} \frac{1}{x^n} = 0 \) for every \( n > 0 \), we obtain:
\[ = \frac{5 \cdot 0 + 7 \cdot 0}{2 + 4 \cdot 0 + 0} = \frac{0}{2} = 0. \]
Result
\[ \lim_{x \to \infty} \frac{5x^3 + 7}{2x^5 + 4x^2 + 1} = 0. \]
This exercise shows how the notable limit is useful for analyzing the behavior of rational functions when \( x \to \infty \).