First slide

Continuity

Question

The value of $f (2)$ so that $f (x)$ $= \frac{2^{x + 2} - 16}{4^{x} - 16}$ is continuous at $x = 2$ is

Difficult

A

2
B

$\frac{1}{2}$
C

-2
D

$- \frac{1}{2}$

Solution

$f (2) = \lim_{x \to 2} f (x) = \lim_{x \to 2} \frac{2^{x + 2} - 16}{4^{x} - 16} = \lim_{x \to 2} \frac{{4.2}^{x} . \log 2}{4^{x} . \log 4} = \frac{{4.2}^{2} . \log 2}{4^{2} .2 \log 2} = \frac{1}{2}$ .

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