Question

Assume that \(\lim _{\theta \rightarrow 1} \mathrm{f}(\theta)\) exists and $\frac{\theta^{2}+\theta-2}{\theta+3} \leq \frac{\mathrm{f}(\theta)}{\theta^{2}} \leq \frac{\theta^{2}+2 \theta-1}{\theta+3}$ holds for certain interval containing the point \(\theta=-1\) then $\lim _{\theta \rightarrow 1} \mathrm{f}(\theta)$ (A) is equal to \(\mathrm{f}(-1)\) (B) is equal to 1 (C) is non-existent (D) is equal to \(-1\)

Short answer

Short Answer: The limit of f(θ) as θ approaches 1 lies between 0 and 1/2, i.e., \(0 \leq \lim_{\theta \rightarrow 1} \mathrm{f}(\theta) \leq \frac{1}{2}\). However, none of the given options (A, B, C, or D) match this result.

Step-by-step solution

Take the limit of the inequality

First, let's compute the limit as \(\theta\) approaches 1. \(\lim_{\theta \rightarrow 1} \frac{\theta^{2}+\theta-2}{\theta+3} \leq \lim_{\theta \rightarrow 1} \frac{\mathrm{f}(\theta)}{\theta^{2}} \leq \lim_{\theta \rightarrow 1} \frac{\theta^{2}+2 \theta-1}{\theta+3}\)

Apply the limit definition

Applying the limit definition, we get: \(\lim_{\theta \rightarrow 1} \frac{\theta^{2}+\theta-2}{\theta+3} = \frac{1^2+1-2}{1+3} = \frac{0}{4} = 0\) \(\lim_{\theta \rightarrow 1} \frac{\theta^{2}+2 \theta-1}{\theta+3} = \frac{1^2+2\cdot1-1}{1+3} = \frac{2}{4} = \frac{1}{2}\) So the inequality becomes: \(0 \leq \lim_{\theta \rightarrow 1} \frac{\mathrm{f}(\theta)}{\theta^{2}} \leq \frac{1}{2}\)

Multiply the inequality by \(\theta^2\)

Now, multiply the inequality by \(\theta^2\): \(0 \cdot \theta^{2} \leq \mathrm{f}(\theta) \leq \frac{1}{2} \cdot \theta^{2}\) As \(\lim_{\theta \rightarrow 1} \mathrm{f}(\theta)\) exists, we can write: \(\lim_{\theta \rightarrow 1} 0 \cdot \theta^{2} \leq \lim_{\theta \rightarrow 1} \mathrm{f}(\theta) \leq \lim_{\theta \rightarrow 1} \frac{1}{2} \cdot \theta^{2}\)

Simplify the inequality

Simplify: \(0 \leq \lim_{\theta \rightarrow 1} \mathrm{f}(\theta) \leq \frac{1}{2}\) We can see that the limit lies between 0 and 1/2. Therefore, we can eliminate options A, B, and D. The only option left is (C), which states the limit is non-existent, but this contradicts our assumption that the limit exists. So, it seems there is a mismatch in the options or the question. It is important to verify if the given question or options are correct.