First slide

Continuity

Question

$Let f (x) = \frac{(256 + a x)^{\frac{1}{8}} - 2}{(32 + b x)^{\frac{1}{5}} - 2} . If f is continuous at x = 0 then the value of \frac{a}{b} is$

Moderate

A

$\frac{8}{5} f (0)$
B

$\frac{32}{5} f (0)$
C

$\frac{64}{5} f (0)$
D

$\frac{16}{5} f (0)$

Solution

$f (0) = \underset{x \to 0}{L t} \frac{{(256 + a x)}^{\frac{1}{8}} - 2}{{(32 + b x)}^{\frac{1}{5}} - 2}$
Use L – Hospital Rule
$\begin{array}{l} = \underset{x \to 0}{L t} \frac{\frac{1}{8} {(256 + a x)}^{\frac{- 7}{8}} . a}{\frac{1}{5} {(32 + b x)}^{\frac{- 4}{5}} . b} \\ ∴ \frac{a}{b} = \frac{64}{5} f (0) \end{array}$

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