First slide

Continuity

Question

The value of $f (0)$ so that $f (x) = \frac{\sqrt{4 + 3 x} - 2}{x}$ is continuous at $x = 0$ is

Easy

A

$- \frac{1}{2}$
B

$\frac{1}{2}$
C

$\frac{3}{4}$
D

$- \frac{3}{4}$

Solution

$f (0) = \underset{x \to 0}{L t} f (x) = \lim_{x \to 0} \frac{\sqrt{4 + 3 x} - 2}{x} = \underset{x \to 0}{L t} \frac{4 + 3 x - 4}{x [\sqrt{4 + 3 x} + 2]} = \frac{3}{4}$

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