Question

Assume that ${lim}_{\theta \to -1}f\left(\theta \right)$ exists and $\frac{{\theta }^2+\theta -2}{\theta +3}\le \frac{f\left(\theta \right)}{{\theta }^2}\le \frac{{\theta }^2+2\theta -1}{\theta +3}$ holds for certain interval containing the point $\theta =-1$ then ${lim}_{\theta \to -1}f\left(\theta \right)$

• A. is equal to $f(-1)$
• B. is equal to $1$
• C. is non existent
• D. is equal to $-1$

Answer 1

The correct option is D is equal to $-1$

$\underset{\theta \to -1}{lim}\frac{{\theta }^2+\theta -2}{\theta +1}=-1=\underset{\theta \to -1}{lim}\frac{{\theta }^2+2\theta -1}{\theta +3}$
Therefore
$\underset{\theta \to -1}{lim}\frac{f\left(\theta \right)}{{\theta }^2}=\underset{\theta \to -1}{lim}f\left(\theta \right)=-1$
Hence, option 'D' is correct.