Question

Prove the given limit using an \(\varepsilon-\delta\) proof. \(\lim _{x \rightarrow 0} \sin x=0\) (Hint: use the fact that \(|\sin x| \leq|x|,\) with equality only when \(x=0 .)\)

Short answer

The limit \(\lim_{x \rightarrow 0} \sin x = 0\) is proven using \(\varepsilon-\delta\) by choosing \(\delta = \varepsilon\).

Step-by-step solution

Understanding the Epsilon-Delta Definition of a Limit

To prove the limit \( \lim _{x \rightarrow 0} \sin x=0 \) using an \(\varepsilon-\delta\) proof, we start by stating that for any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(|x-0| < \delta\), then \(|\sin x - 0| < \varepsilon\). This means we want \(|\sin x| < \varepsilon\) whenever \(|x| < \delta\).

Using the Sine Inequality

We are given the inequality \(|\sin x| \leq |x|\). This means that if \(|x| < \delta\), then \(|\sin x| \leq |x| < \delta\). By choosing \(\delta\) such that \(\delta = \varepsilon\), we ensure that \(|\sin x| < \varepsilon\) whenever \(|x| < \delta\).

Setting the Delta Condition

Choose \(\delta = \varepsilon\). By applying this choice, whenever \(|x| < \delta\), we have \(|\sin x| \leq |x| < \delta = \varepsilon\). This guarantees that \(|\sin x| < \varepsilon\), which satisfies the \(\varepsilon-\delta\) condition for the limit.

Conclusion of the Proof

Since for any \(\varepsilon > 0\), we can find a \(\delta = \varepsilon\) such that if \(|x| < \delta\), then \(|\sin x| < \varepsilon\), we have shown through an \(\varepsilon-\delta\) proof that \(\lim _{x \rightarrow 0} \sin x=0 \). This demonstrates that the limit is true.

Key concepts

Sine Inequality

The sine inequality plays an important role when proving limits involving the sine function. It states that for any angle

• \(|\sin x| \leq |x|\)

This inequality helps us understand the behavior of the sine function as it approaches zero. Essentially, it tells us that the absolute value of \(\sin x\) is always less than or equal to the absolute value of \(x\). This is crucial because, near zero, the sine function behaves similarly to the line \(y=x\).
One way to visualize this is by imagining the graph of \(\sin x\) close to the origin. You'll notice that it hugs the line \(y=x\) very closely. This fact makes the sine inequality a handy tool in calculus, especially when working with limits.
So, when working with limits like \(\lim_{x \to 0} \sin x = 0\), the sine inequality provides the foundational understanding that helps confirm the behavior of the sine function in terms of its approximations.

Limit of Sine Function

The limit of the sine function as the input approaches zero is a fundamental concept in calculus. Formally, we express it as

• \(\lim_{x \to 0} \sin x = 0\)

This means that as \(x\) gets closer and closer to zero, \(\sin x\) gets closer to zero as well. It's essential to grasp this not only for theoretical proofs like the \(\varepsilon-\delta\) proof but also for practical applications in calculus.
Practically speaking, when you input values of \(x\) that are very small, \(\sin x\) will also yield a very small result, demonstrating the function's continuity and smooth transition to zero at the origin.
In calculus, understanding the limit of \(\sin x\) helps in evaluating derivatives, integrals, and in solving differential equations. It lays the groundwork for trigonometric identities and their applications in various mathematical problems.

Calculus Limit Proof

A calculus limit proof, particularly the \(\varepsilon-\delta\) proof, is a rigorous way to demonstrate that a function approaches a particular value as its input gets infinitely close to a given point. Let's break down how this proof works for proving \(\lim_{x \to 0} \sin x = 0\).

• Start by understanding that we want \(|\sin x| < \varepsilon\) whenever \(|x| < \delta\).
• Based on the sine inequality \(|\sin x| \leq |x|\), we choose \(\delta = \varepsilon\). This exploits the inequality to ensure \(\sin x\) approaches zero.
• This means, if \(|x| < \delta\), it follows \(|\sin x| < \varepsilon\), as desired for proving the limit.

This method guarantees rigor in calculus by ensuring the function behaves as we predict. The \(\varepsilon-\delta\) approach not only shows that \(\sin x\) approaches zero but also provides the exact quantification of how close \(x\) must be to zero for \(\sin x\) to be within any desired tiny range from zero.
The beauty of this proof lies in its precision and its capacity to handle logical conclusions from first principles, explaining why calculus earns its reputation for mathematical rigor.