Question

If $7-\frac{x^2}{12}\le f\left(x\right)\le 7+\frac{x^3}{5}$ for all $x\ne 0$, then $\underset{x\to 0}{lim}f\left(x\right)=$

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$7$

$TakeLHSf\left(x\right)=\begin{array}{c}lim\\ x\to 0^{-}\end{array}7-\frac{{(0-h)}^2}{12}(whereh\to 0)$
$=7$
$TakeRHSf\left(x\right)=\begin{array}{c}lim\\ x\to 0^{+}\end{array}7+\frac{{(0-h)}^3}{5}(whereh\to 0)$
$=7$
RHS=LHS=7