If lim _{x rightarrow 0} frac{ax-(e^{4x}-1)}{ | vedclass

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(C) Given the limit $\lim _{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)} = b$. Since the limit exists,the form must be $\frac{0}{0}$ at $x=0$.

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the limit $\lim _{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)} = b$. Since the limit exists,the form must be $\frac{0}{0}$ at $x=0$. Using the expansion $e^{4x} = 1 + 4x + \frac{(4x)^2}{2!} + \dots$,we have: $\lim _{x \rightarrow 0} \frac{ax - (1 + 4x + 8x^2 + \dots - 1)}{ax(4x + 8x^2 + \dots)} = \lim _{x \rightarrow 0} \frac{(a-4)x - 8x^2}{4ax^2 + 8ax^3}$. For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $a-4 = 0$,which gives $a = 4$. Now,substituting $a=4$ into the limit: $b = \lim _{x \rightarrow 0} \frac{-8x^2}{4(4)x^2 + 8(4)x^3} = \lim _{x \rightarrow 0} \frac{-8x^2}{16x^2 + 32x^3} = \lim _{x \rightarrow 0} \frac{-8}{16 + 32x} = -\frac{8}{16} = -\frac{1}{2}$. Thus,$a = 4$ and $b = -\frac{1}{2}$. Finally,$a - 2b = 4 - 2(-\frac{1}{2}) = 4 + 1 = 5$.

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 ....... .

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