91Ó°ÊÓ

Calculating limits is a fundamental aspect of calculus, which helps us understand the behavior of functions as they approach specific values. In simple terms, a limit evaluates what value a function approaches as the input gets nearer to a particular point. This becomes especially crucial when direct substitution doesn't provide a straightforward answer, such as at points of discontinuity or singularity.

The exercise we are examining involves computing the limit of a difference of fractions, which inherently can be tricky due to potential points of singularity. To handle this, it is essential to simplify the expression first. For this problem, knowing that \[ x^2 - 1 = (x-1)(x+1) \]is essential as it allows us to combine fractions and better understand the function's behavior as \(x\) approaches 1. By simplifying the expression \[ \frac{1}{x-1} - \frac{a}{x^2-1} \]to \[ \frac{(x+1) - a}{(x-1)(x+1)} \],we set the stage for successfully calculating the limit.

Singularity in limits occurs where a function does not have a well-defined output at certain points, typically due to division by zero or other undefined expressions. In the problem, the singularity arises because both \( x-1 \) and \( x^2-1 \) incorporate factors that cancel out, causing potential undefined behavior at \( x = 1 \).

Recognizing singular points is crucial in limit problems. These are often where a function behaves unpredictably, requiring careful analysis. Here, simplifying the expression gives us insights into how to handle the singularity. Essentially, the singularity at \( x = 1 \) can potentially be removed if the numerator also becomes zero at that point, leading to a finite limit.

In our scenario, setting \( a = 2 \) ensures that the numerator becomes zero when \( x = 1 \):\[ 2 - a = 0 \] This balances the equation, allowing for cancellation of the undefined parts and yielding a meaningful limit.

L'Hopital's Rule is a valuable tool in calculus for determining limits that initially appear as indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if given a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and denominator separately and then take the limit again.

In our exercise, once we have simplified and reached the expression \[ \frac{0}{(x-1)(x+1)} \], the use of L'Hopital's Rule confirms the limit calculation. By differentiating the components and reevaluating the limit, we can ascertain whether the function approaches a finite value as \( x \) nears 1.

For this problem, differentiation of \( 2 - x \) and \((x-1)(x+1)\) aids in clearing the indeterminate form. Although L'Hopital's Rule can simplify many singular points, here, understanding how \( a = 2 \) handles the numerical zeroing is vital, leading to a solution of \[ \frac{1}{2} \]​​when \( x \) approaches 1.