Question

Does x Reach a in Limit Definition?

Original question: We know that a limit is the value that f(x) approches as x gets closer and clsoer to some vlaue a, but does the x actually reach a or it just keeps getting close without reaching it?

Expert Verified Solution

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Answer

In the definition of a limit, xx approaches the value aa but does not need to actually reach or equal aa; the process involves xx getting arbitrarily close to aa from values nearby, without requiring x=ax = a. This ensures the limit focuses on the function's behavior in the vicinity of aa, regardless of whether the function is defined or continuous at exactly x=ax = a.

Explanation

  1. Recall the formal definition of a limit
    The limit limxaf(x)=L\lim_{x \to a} f(x) = L means that for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that if 0<xa<δ0 < |x - a| < \delta, then f(x)L<ϵ|f(x) - L| < \epsilon.
    This definition emphasizes that xx is restricted to 0<xa<δ0 < |x - a| < \delta, excluding x=ax = a itself, so xx never actually reaches aa in the evaluation— it only approaches it. This setup allows us to study the trend near aa without depending on the value at aa.

  2. Understand why x=ax = a is excluded
    By excluding x=ax = a, the limit avoids issues like discontinuities or undefined points at aa; it purely describes how f(x)f(x) behaves as xx nears aa from either side (or both).
    This is pedagogically important because it separates the limit (a directional approach) from the function's value at the point, enabling analysis even when f(a)f(a) doesn't exist or differs from the limit.

  3. Apply to an example
    Consider f(x)=x21x1f(x) = \frac{x^2 - 1}{x - 1}, which simplifies to x+1x + 1 for x1x \neq 1, so limx1f(x)=2\lim_{x \to 1} f(x) = 2.
    Here, f(1)f(1) is undefined (division by zero), but the limit exists because as xx gets close to 1 (e.g., x=0.999x = 0.999 or x=1.001x = 1.001), f(x)f(x) approaches 2—showing xx doesn't need to reach 1 for the limit to be determined.

Final Answer

The value of xx does not actually reach aa in the limit definition; it only approaches aa arbitrarily closely.
xapproachesabutdoesnotneedtoequala\boxed{x approaches a but does not need to equal a}

Common Mistakes

  • Confusing the limit with the function value at aa: Students often assume limxaf(x)\lim_{x \to a} f(x) requires f(a)f(a) to exist and equal the limit, but the limit can exist even if f(a)f(a) is undefined or different.
  • Thinking the limit process includes x=ax = a: Many overlook the 0<xa0 < |x - a| condition, leading to errors when the function is discontinuous at aa.

FAQ

What does the formal limit definition say about x reaching a?

The limit as x approaches a equals L if for every epsilon > 0, there's a delta > 0 such that 0 < |x - a| < delta implies |f(x) - L| < epsilon, excluding x = a.

Why is x = a excluded from the limit process?

Excluding x = a allows the limit to focus on behavior near a, even if f(a) is undefined or discontinuous, separating the approach from the value at the point.

Can you give an example where the limit exists but f(a) is undefined?

For f(x) = (x² - 1)/(x - 1), which simplifies to x + 1 for x ≠ 1, the limit as x approaches 1 is 2, though f(1) is undefined due to division by zero.

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