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L'Hospital Rule to Remove Indeterminate Form

If lim x → 0 ...

Question

If $lim x \to 0 \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)}$ exists and is equal to $b,$ then the value of $a - 2 b$ is

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Solution

$lim x \to 0 \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)}$
Applying L'Hospital Rule
$lim x \to 0 \frac{a - 4 e^{4 x}}{a (e^{4 x} - 1) + a x (4 e^{4 x})}$
So for limit to exist, $a = 4$
Applying L′Hospital Rule
$lim x \to 0 \frac{- 16 e^{4 x}}{a (4 e^{4 x}) + a (4 e^{4 x}) + a x (16 e^{4 x})}$
$= \frac{- 16}{4 a + 4 a} = \frac{- 16}{32} = - \frac{1}{2} = b$
$∴ a - 2 b = 4 - 2 (\frac{- 1}{2}) = 4 + 1 = 5$

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L'Hospital Rule to Remove Indeterminate Form

Standard XII Mathematics

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