Using the Formal Definition

Prove the limit statements in Exercises 37–50.

lim x→0 x² sin (1/x) = 0

<IMAGE>

검증된 단계별 안내

Understand the formal definition of a limit: For a function f(x) to have a limit L as x approaches a, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε.

Identify the function and the limit: Here, f(x) = x² sin(1/x) and we want to prove that lim x→0 f(x) = 0.

Set up the inequality using the formal definition: We need to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x| < δ, then |x² sin(1/x) - 0| < ε.

Simplify the expression: Notice that |x² sin(1/x)| = |x²| |sin(1/x)|. Since |sin(1/x)| ≤ 1 for all x, we have |x² sin(1/x)| ≤ |x²|.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limit Definition

The formal definition of a limit states that for a function f(x), the limit as x approaches a value c is L if, for every ε > 0, there exists a δ > 0 such that whenever 0 < |x - c| < δ, it follows that |f(x) - L| < ε. This definition is crucial for proving limit statements rigorously.

Squeeze Theorem

The Squeeze Theorem is a method used to find limits of functions that are difficult to evaluate directly. It states that if f(x) ≤ g(x) ≤ h(x) for all x near c (except possibly at c), and if the limits of f(x) and h(x) as x approaches c are both L, then the limit of g(x) as x approaches c is also L. This theorem is particularly useful for functions involving oscillatory behavior, like sin(1/x).

Behavior of Oscillatory Functions

Oscillatory functions, such as sin(1/x), do not settle at a single value as x approaches 0; instead, they oscillate between -1 and 1. Understanding this behavior is essential when evaluating limits involving such functions, as it allows us to analyze their impact on the overall limit, particularly when multiplied by terms that approach zero, like x² in this case.

Using the Formal DefinitionProve the limit statements in Exercises 37–50.lim x→0 x² sin (1/x) = 0<IMAGE>