91Ó°ÊÓ

When we analyze the behavior of a function as it approaches a particular point, we're trying to understand how it acts near that point without necessarily evaluating it at the point itself. For example, the function \( f(x) = \frac{-1}{x^2(x+1)} \) becomes more complicated as \( x \to 0 \). While the numerator remains constant at \(-1\), the denominator, \(x^2(x+1)\), has a significant influence on the behavior of the function.

• The numerator value \(-1\) suggests that regardless of the denominator's sign, the overall fraction will always be negative as we approach zero.
• As \( x \to 0 \), \(x^2\) gets very small because it involves squaring a small number, and \(x+1\) will hover slightly above or below 1 depending on whether \(x\) is positive or negative.

The combination of these factors means as \(x\) approaches 0 from either side, the function's value continues to drop significantly due to division by a small positive number, leading to a trend toward \(-\infty\). Understanding this contextual behavior is critical when considering limits, particularly in complex functions.

To find the right-hand limit, we analyze what happens as \( x \) approaches \( 0 \) from the positive side (\( x \to 0^+ \)). When we plug in values closer and closer to zero but remain positive, we observe:

• \( x^2 \), being the square of a positive number, is still positive and close to \(0\).
• \( x+1 \) slightly exceeds \(1\) as \(x\) is positive, bringing the denominator to be close to \(0^+\).

Consequently, the denominator, although a small positive value, results in an extremely large negative quotient given the \(-1\) in the numerator. Hence, as \( x \to 0^+ \), the function value trends towards \(-\infty\). This indicates that the right-hand limit of the function as \( x \to 0 \) is \(-\infty\).

Understanding the left-hand limit involves examining where \( x \) approaches 0 from the negative side (\( x \to 0^- \)). This means we consider slightly negative \( x \) values.

• Despite \( x \) being negative, \( x^2 \) remains a positive number due to squaring, still approaching \(0\).
• \( x+1 \) now becomes slightly less than \(1\) since the negative \(x\) diminishes the sum.

The combination signifies that the denominator again inclines toward a very small positive number, which results in profoundly negative function values because of the \(-1\) on top. As \( x \to 0^- \), the function dips towards \(-\infty\), confirming that the left-hand limit on this side is also \(-\infty\). Hence, both sides match up, signifying this limit is definitive.