Question

If the function $f\left(x\right)=\{\begin{array}{ccc}\frac{2^{x+2}-16}{4^x-16}& & forx\ne 2\\ A& & x=2\end{array}$ is continuous at $x=2$, then $A=$

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $0$

Answer 1

The correct option is B $\frac{1}{2}$

Since, the function is continuous at $x=2$
$A=\underset{x\to 2}{lim}f\left(x\right)$
Since, on applying the limit this is of the $\frac{0}{0}$ form, we apply L-Hospital's Rule
$\Rightarrow \underset{x\to 2}{lim}\frac{4.2^{x}log2}{4^{x}log4}$
Now applying the limit,
$A=\frac{1}{2}$