91Ó°ÊÓ

Factoring polynomial expressions is a crucial skill in calculus, especially when dealing with limits. In our exercise, we have the expression \( \frac{x^2 - 3x + 2}{x^3 - 4x} \). Breaking this down requires us to find common factors in both the numerator and the denominator.
The numerator, \( x^2 - 3x + 2 \), can be factored into \((x-1)(x-2)\). Meanwhile, the denominator \( x^3 - 4x \) factors into \( x(x^2 - 4) = x(x-2)(x+2) \). This factoring simplifies our expression considerably.
Once factored, the expression becomes:

• Numerator: \((x-1)(x-2)\)
• Denominator: \(x(x-2)(x+2)\)

Cancel the common factor \(x-2\), which exists in both the numerator and the denominator, provided \(x eq 2\). This results in a new simpler expression: \( \frac{x-1}{x(x+2)} \). Factoring is essential for simplifying expressions and solving limits, as demonstrated in this context.

When evaluating limits, it's important to understand how the value of \(x\) affects the expression as \(x\) approaches particular points. The expression \( \frac{x-1}{x(x+2)} \) reveals different behaviors for different limits.
For instance, as \( x \to 2^+ \), the expression evaluates to \( \frac{1}{8} \). This is straightforward because the cancellation and subsequent substitution give a real number.
Now, when considering \( x \to -2^+ \), both terms \(x\) and \(x+2\) trend towards zero but on opposite spectrums (negative and positive, respectively). This causes implications of behavior approaching infinity, but with a negative sign indicating \( -\infty \).
Finally, analyzing \( x \to 1^+ \), we note \( x-1 \) turns to zero, making the entire fraction tend towards 0. It is crucial to identify these trends in values when determining a limit for different cases.

In calculus, an indeterminate form can often be a roadblock in calculating limits. It occurs when direct substitution into a limit expression leads to forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), which provide no immediate answer.
Our expression rises potentially indeterminate forms when \( x-2 \) was not considered in the factorization step. Originally, substitution into \( x \) without cancelling common factors results in \( \frac{0}{0} \) at particular points like \( x = 2 \). By factoring and canceling properly, we resolve these indeterminate forms to something manageable: \( \frac{x-1}{x(x+2)} \).
This requires careful manipulation, such as factoring and cancelation as shown here, which prevent the calculation from getting stuck in the indeterminate stages.

Infinity often appears in limits as a result of division by zero or other behaviors where values grow without bound. In our expression, infinity surfaces under two circumstances.
For \( x \to -2^+ \), the denominator \( x(x+2) \) reaches zero while the negative multiplication indicates \( -\infty \). This happens because \( x \) approaches \(-2\) but never reaches it, sending the fraction numerator's dominance to -\(\infty\).
Similarly, as \( x \to 0^- \), the denominator approaches zero rapidly because the \( x \) term directly influences it. Here, \( \frac{-1}{x \cdot 2} \) takes over, spiraling the whole fraction's behavior towards \(-\infty\) again. These cases demonstrate how infinity can result in limits when functions demonstrate unbounded growth or shrinkage near specific x-values.