Problem 13 - The Calculus with Analytic Geometry

Question

Find the limits, and when applicable indicate the limit theorems being used. $$ \lim _{y \rightarrow+\infty} \frac{2 y^{3}-4}{5 y+3} $$

Step-by-Step Solution

Answer

The limit is \(\rightarrow\infty\).

Step 1 - Identify the leading terms in the numerator and denominator

In the given function \(\frac{2 y^{3}-4}{5 y+3} \), the leading term in the numerator is \(2y^3\) and in the denominator is \(5y\). These terms dominate the behavior of the function as \( y \to +\infty \).

Step 2 - Simplify the expression by factoring out the leading terms

Rewrite the limit expression by factoring out the leading terms: \[ \frac{2 y^{3}-4}{5 y+3} = \frac{2 y^{3}(1 - \frac{2}{y^3})}{5 y(1 + \frac{3}{5 y})} \]

Step 3 - Cancel out common factors

Cancel out the common factor \(y\): \[ \frac{2 y^{3}(1 - \frac{2}{y^3})}{5 y(1 + \frac{3}{5 y})} = \frac{2 y^2 (1 - \frac{2}{y^3})}{5 (1 + \frac{3}{5 y})} \]

Step 4 - Evaluate the limit as \( y \to +\infty \)

As \( y \to +\infty, \) the terms \( \frac{-2}{y^3} \) and \( \frac{3}{5y} \) approach 0. Thus, the simplified limit is: \[ \frac{2 y^2 (1 - 0)}{5 (1 + 0)} = \frac{2 y^2}{5} \].

Step 5 - Determine the final simplified form

Finally, as \( y \to +\infty, \) \(2 y^2 \) becomes unbounded, indicating the expression grows towards \(\frac{2 y^2}{5} = y \to +\infty\). This indicates the original limit is \(\rightarrow\infty\).

Key Concepts

Limit Evaluation

In calculus, evaluating limits helps us understand the behavior of functions as they approach specific points or infinity. For this exercise, we need to find the limit as y approaches positive infinity. We focus on the behavior of the function by examining the highest degree terms, as these will dominate the function's growth. This method simplifies our calculations and helps us accurately determine the limit.

Leading Terms

Identifying the leading terms in both the numerator and the denominator of a function is crucial when evaluating limits at infinity. Here, the given function is \(\frac{2 y^3 - 4}{5 y + 3}\).
- For the numerator: The leading term is 2y³ since it's of the highest degree.
- For the denominator: The leading term is 5y, which also has the highest degree (linear term). The behavior as y approaches infinity will mainly be influenced by these leading terms.

Factoring

Factoring out leading terms is a powerful technique to simplify expressions.
We start with \(\frac{2 y^3 - 4}{5 y + 3}\).
Factoring out the leading terms from both the numerator and denominator:
\(\frac{2 y^3 (1 - \frac{2}{y^3})}{5 y(1 + \frac{3}{5 y})}\).
This helps us ignore smaller terms as they become insignificant when y approaches infinity. Simplifying further, cancel out the common factor y:
\(\frac{2 y^2 (1 - \frac{2}{y^3})}{5 (1 + \frac{3}{5 y})}\).

Infinity Behavior

Understanding how functions behave as their variables approach infinity is key. In our example, as y approaches positive infinity:
- The terms approximating fractions (like \( \frac{-2}{y^3} \) and \( \frac{3}{5y} \)) approach zero.
Substituting these approximations, we get:
\(\frac{2 y^2 (1 - 0)}{5 (1 + 0)} = \frac{2 y^2}{5}\).
As y continues to increase, \( 2 y^2 \) becomes unbounded, implying the limit grows towards infinity. Therefore, the limit is \rightarrow \infty.