Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limh→0 sin(sin h) / sin h

검증된 단계별 안내

First, recognize that the problem involves finding the limit of the expression lim(h→0) sin(sin(h)) / sin(h). This is a limit problem that can be approached using the known limit lim(θ→0) sin(θ) / θ = 1.

To solve this, consider the substitution u = sin(h). As h approaches 0, u also approaches 0 because sin(h) is continuous and sin(0) = 0.

Rewrite the original limit in terms of u: lim(u→0) sin(u) / u. This transformation is valid because as h approaches 0, u approaches 0.

Now, apply the known limit property: lim(θ→0) sin(θ) / θ = 1. Since u is approaching 0, this property can be directly applied to the expression lim(u→0) sin(u) / u.

Conclude that the limit of the original expression lim(h→0) sin(sin(h)) / sin(h) is equal to 1, based on the substitution and the known limit property.

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주요 개념

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

Limit of a Function

The limit of a function describes the behavior of the function as the input approaches a particular value. In calculus, understanding limits is crucial for analyzing the continuity and differentiability of functions. For the given problem, evaluating the limit as h approaches 0 is essential to determine the behavior of the function sin(sin h) / sin h.

Squeeze Theorem

The Squeeze Theorem is a method used to find the limit of a function by 'squeezing' it between two other functions whose limits are known and equal. This theorem is particularly useful when direct substitution in a limit results in an indeterminate form. In this problem, the Squeeze Theorem can help confirm the limit of sin(sin h) / sin h as h approaches 0.

Limit of sin(θ)/θ as θ approaches 0

The limit limθ→0 sin(θ)/θ = 1 is a fundamental result in calculus, often used to evaluate limits involving trigonometric functions. This limit is derived from the unit circle and the small-angle approximation. In the given problem, this concept is directly applied to simplify and solve the limit of sin(sin h) / sin h as h approaches 0.

Using limθ→0 sin θ / θ = 1Find the limits in Exercises 23–46.limh→0 sin(sin h) / sin h