Question

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

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

Answer 1

Answer: (A), (D)

The correct options are
A is equal to $f(-1)$
D is equal to $-1$

${lim}_{x\to -1}\frac{x^2+x-2}{x+3}=-1={lim}_{x\to -1}\frac{x^2+2x-1}{x+3}$
$\therefore {lim}_{x\to -1}\frac{f\left(x\right)}{x^2}={lim}_{x\to -1}f\left(x\right)=-1$

Which is also equal to$f(-1)$
Hence, options 'A' and 'D' are correct.