Problem 263 - Calculus Volume 1

Question

For the following exercises, evaluate the limit. $$ \lim _{x \rightarrow \infty} \frac{x^{2}-2 x+5}{x+2} $$

Step-by-Step Solution

Answer

The limit is \( \infty \).

Step 1: Identify the Limit Type

We need to evaluate \( \lim _{x \rightarrow \infty} \frac{x^{2}-2x+5}{x+2} \). This is a limit as \( x \to \infty \), which means we are looking at the behavior of the function as \( x \) becomes very large.

Step 2: Divide by the Highest Power of x

The highest power of \( x \) in the expression is \( x^2 \). To simplify the fraction, divide both the numerator and the denominator by \( x \), the highest power of \( x \) found in the denominator:\[\lim _{x \rightarrow \infty} \frac{x^{2}-2x+5}{x+2} = \lim _{x \rightarrow \infty} \frac{x^2/x - 2x/x + 5/x}{x/x + 2/x}\] which simplifies to:\[\lim _{x \rightarrow \infty} \frac{x - 2 + \frac{5}{x}}{1 + \frac{2}{x}}\]

Step 3: Evaluate the Limit as x approaches Infinity

As \( x \) approaches infinity, any term with \( \frac{1}{x} \) in it will approach 0. Therefore, the expression simplifies to:\[\lim _{x \rightarrow \infty} \frac{x - 2 + 0}{1 + 0} = \lim _{x \rightarrow \infty} (x - 2)\]As \( x \to \infty \), \( x - 2 \to \infty \).

Step 4: Conclude the Result

The value of \( \lim _{x \rightarrow \infty} \frac{x^{2}-2x+5}{x+2} = \infty \). This means the limit does not exist in finite terms, but instead grows indefinitely large.

Key Concepts

Limit at Infinity

When we talk about the "limit at infinity," we are exploring what happens to a function as the variable approaches infinity. In simpler terms, we want to understand how a function behaves when the value of the variable becomes very large.
For example, in the exercise, we evaluated \(lim _{x \rightarrow \infty} \frac{x^{2}-2 x+5}{x+2}\). Here, we are observing how the function \(\frac{x^{2}-2 x+5}{x+2}\) changes as \(x\) becomes larger and larger.

• As \(x\) approaches infinity, we pay close attention to the dominant terms in the expression. The dominant terms are those with the highest powers of \(x\).
• In the numerator, \(x^2\) dominates because it grows faster than \(-2x + 5\).
• In the denominator, \(x\) is the dominant term.

By focusing on these dominant terms, we can simplify the expression and determine its behavior as \(x\) increases without specifically needing to calculate infinite values.

Polynomial Division

Polynomial division helps us simplify and evaluate limits more easily. When we are dealing with rational functions like \(\frac{x^{2}-2 x+5}{x+2}\), dividing both the numerator and the denominator by the highest power of \(x\) in the denominator can make the limit much clearer.
In this problem, the highest power in the denominator is \(x\). By dividing each term by \(x\), we achieve the following simplified form:
\[\frac{x - 2 + \frac{5}{x}}{1 + \frac{2}{x}}\]

• Each term with \(1/x\) reduces significantly as \(x\) approaches infinity, simplifying our analysis.
• Only terms involving high degrees of \(x\) remain significant when assessing behavior at infinity.

This process unlocks the dominant behavior of the function, which mirrors the highest degree terms, providing a simpler expression to evaluate the limit.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches a specific point or as the variable heads to infinity. It gives insight into the end behavior of a polynomial or rational function.
In our example, as \(x\) approaches infinity, the expression \(\frac{x - 2 + 0}{1 + 0}\) simplifies to \(x - 2\), which also approaches infinity. This implies a certain asymptotic behavior:

• The function \(\frac{x^{2}-2 x+5}{x+2}\) behaves like a line as \(x\) becomes very large because the dominant term in the simplified version is \(x\).
• The lack of horizontal or slant asymptote in this case implicates that the graph of the function will keep increasing without leveling off.

The asymptotic behavior in such cases often means the original function becomes unbounded, either heading to infinity or negative infinity as \(x\) grows.