Questions: Determine the following limit. lim θ→∞ (sin 9θ)/(11θ)

Solution Steps

To determine the limit of \(\lim _{\theta \rightarrow \infty} \frac{\sin 9 \theta}{11 \theta}\), we can use the fact that the sine function oscillates between -1 and 1. As \(\theta\) approaches infinity, the denominator \(11\theta\) grows without bound, causing the fraction to approach 0.

We want to evaluate the limit: \[ \lim _{\theta \rightarrow \infty} \frac{\sin(9\theta)}{11\theta} \]

The sine function, \(\sin(9\theta)\), oscillates between -1 and 1 for all values of \(\theta\). Therefore, we can express the bounds of the function as: \[ -1 \leq \sin(9\theta) \leq 1 \]

As \(\theta\) approaches infinity, the denominator \(11\theta\) grows without bound. Thus, we can analyze the limit: \[ \frac{-1}{11\theta} \leq \frac{\sin(9\theta)}{11\theta} \leq \frac{1}{11\theta} \] As \(\theta \rightarrow \infty\), both \(\frac{-1}{11\theta}\) and \(\frac{1}{11\theta}\) approach 0.

By the Squeeze Theorem, since \(\frac{\sin(9\theta)}{11\theta}\) is squeezed between two expressions that both approach 0, we conclude that: \[ \lim _{\theta \rightarrow \infty} \frac{\sin(9\theta)}{11\theta} = 0 \]

Final Answer