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If \lim_{x\rightarrow 0}\frac{ax - (e^{4x}-1)}{ax (e^{4x}-1)} exists and is equal to b, then the value of a - 2b is ______.
Option: 1 5
Option: 2 4
Option: 3 3
Option: 4 2

Answers (1)

best_answer

\\\lim _{x \rightarrow 0} \frac{a x-\left(e^{4 x}-1\right)}{\operatorname{ax}\left(e^{4 x}-1\right)}\\\text{this is in }\frac{0}{0}\;\text{form}

\\\lim _{x \rightarrow 0} \frac{\left (a x-\left(e^{4 x}-1\right) \right )4x}{\operatorname{ax}\left(e^{4 x}-1\right)4x}

=\lim _{x \rightarrow 0} \frac{\operatorname{ax}-\left(e^{4 x}-1\right)}{a x \cdot 4 x} \quad \left (\text { Use } \lim _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1 \right )

Apply L'Hopital rule

=\lim _{x \rightarrow 0} \frac{a-4 e^{4 x}}{8 a x}\;\;\;\;\;\;\;\;\;\left ( \frac{a-4}{0}\;\text{form} \right )

for the limit to exist, a = 4

\\=\lim _{x \rightarrow 0} \frac{4-4 e^{4 x}}{32 x} \\ =\lim _{x \rightarrow 0} \frac{1-\mathrm{e}^{4 \mathrm{x}}}{8 \mathrm{x}} \quad\left(\frac{0}{0}\right) \\ =\lim _{x \rightarrow 0} \frac{-\mathrm{e}^{4 x} \cdot 4}{8}=-\frac{1}{2} \\\Rightarrow \mathrm{b}=-\frac{1}{2}

\\a-2 b=4-2\left(-\frac{1}{2}\right) \\ =5

 

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himanshu.meshram

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