Question

If $f\left(x\right)=\frac{x^2+6x}{\sin x}$ , then $\underset{x\to 0^{-}}{lim}f\left(x\right)=$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

${lim}_{x\to 0}^{-}f\left(x\right)=\frac{x^2+6x}{\sin x}={lim}_{h\to 0}f\left(h\right)$

$=\frac{h^2+6(0-h)}{\sin (0-h)}={lim}_{h\to 0}f\left(h\right)=\frac{h-6}{-\sin \left(h\right)/h}=\frac{-6}{-1}=6$
Hence, the answer is "C".