#### Assume that $\mathrm{\lim _{\theta \rightarrow-1} f(\theta)\text{exists and }\frac{\theta^2+\theta-2}{\theta+3} \leq \frac{f(\theta)}{\theta^2} \leq \frac{\theta^2+2 \theta-1}{\theta+3}}$ holds for certain interval containing the point $\mathrm{\theta=-1}$, then $\mathrm{\lim _{\theta \rightarrow-1} f(\theta)}$ is equal to  Option: 1 Option: 2 Option: 3 Option: 4

$\mathrm{\frac{\theta^2+\theta-2}{\theta+3} \leq \frac{f(\theta)}{\theta^2} \leq \frac{\theta^2+2 \theta-1}{\theta+3}}$

Putting $\mathrm{\theta=-1,}$ we get

\mathrm{ \begin{aligned} & \frac{1-1-2}{2} \leq f(-1) \leq \frac{1-2-1}{2} \\ & \Rightarrow-1 \leq f(-1) \leq-1 \Rightarrow f(-1)=-1 \end{aligned} }

Also, $\mathrm{\lim _{\theta \rightarrow-1} \frac{\theta^2+\theta-2}{\theta+3}=-1=\lim _{\theta \rightarrow-1} \frac{\theta^2+2 \theta-1}{\theta+3}}$

Using Sandwich theorem,

$\mathrm{ \lim _{\theta \rightarrow-1} \frac{f(\theta)}{\theta^2}=-1 ; \lim _{\theta \rightarrow-1} f(\theta)=-1=f(-1) }$

Hence option 1 is correct.