If limx→0 ax−e4x−1axe4x−1=b exists finitely then 2(a+b)=

limx→0 ax−e4x−1axe4x−1=b exists finitely⇒ limx→0 a−4e4x−14xae4x−1=b exists finitelyThe numerator becomes a-4 when x→0 and the denominator tends to zero. Therefore, if a -4≠0, LHS→∞ whereas RHS is finite. Therefore, a -4 = 0 i.e. a = 4.When a=4 limx→0 a−4e4x4x−1ae4x−1=b⇒ limx→0 1−e4x−14xe4x−1=b ⇒ limx→0 1−1+(4x)2!+(4x)23!+…4x+(4x)22!+(4x)33!+…=b⇒−12=b∴ 2(a+b)=24−12=7

If limx→0 ax−e4x−1axe4x−1=b exists finitely then 2(a+b)=

limx→0 ax−e4x−1axe4x−1=b exists finitely⇒ limx→0 a−4e4x−14xae4x−1=b exists finitelyThe numerator becomes a-4 when x→0 and the denominator tends to zero. Therefore, if a -4≠0, LHS→∞ whereas RHS is finite. Therefore, a -4 = 0 i.e. a = 4.When a=4 limx→0 a−4e4x4x−1ae4x−1=b⇒ limx→0 1−e4x−14xe4x−1=b ⇒ limx→0 1−1+(4x)2!+(4x)23!+…4x+(4x)22!+(4x)33!+…=b⇒−12=b∴ 2(a+b)=24−12=7

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If $lim_{x \to 0} \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)} = b$ exists finitely then $2 (a + b) =$

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Detailed Solution

$lim_{x \to 0} \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)} = b$ exists finitely

$\Rightarrow lim_{x \to 0} \frac{a - 4 (\frac{e^{4 x} - 1}{4 x})}{a (e^{4 x} - 1)} = b$ exists finitely

The numerator becomes $a - 4$ when $x \to 0$ and the denominator tends to zero. Therefore, if $a - 4 \neq 0,$ $L H S \to \infty$ whereas RHS is finite. Therefore, $a - 4 = 0$ i.e. $a = 4 .$

When $a = 4$

$\begin{aligned} lim_{x \to 0} \frac{a - 4 (\frac{e^{4 x}}{4 x} - 1)}{a (e^{4 x} - 1)} = b \\ \Rightarrow lim_{x \to 0} \frac{1 - (\frac{e^{4 x} - 1}{4 x})}{e^{4 x} - 1} = b \end{aligned}$

$\Rightarrow lim_{x \to 0} \frac{1 - (1 + \frac{(4 x)}{2!} + \frac{(4 x)^{2}}{3!} + \dots)}{4 x + \frac{(4 x)^{2}}{2!} + \frac{(4 x)^{3}}{3!} + \dots} = b \Rightarrow - \frac{1}{2} = b$