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  • Assume that <img alt="\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}}"

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

\mathrm{f(-1)}


Option: 2

1


Option: 3

-2


Option: 4

\text{does not exist}


Answers (1)

best_answer

\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.
 

Posted by

shivangi.bhatnagar

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