If Delta(x) = begin{vmatrix} e^x & -1 sin x | vedclass

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(B) Given $\Delta(x) = \begin{vmatrix} e^x & -1 \ \sin x - 1 & 1 \end{vmatrix}$.Expanding the determinant,we get $\Delta(x) = (e^x)(1) - (-1)(\sin x

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$2$

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(B) Given $\Delta(x) = \begin{vmatrix} e^x & -1 \ \sin x - 1 & 1 \end{vmatrix}$.Expanding the determinant,we get $\Delta(x) = (e^x)(1) - (-1)(\sin x - 1)$.$\Delta(x) = e^x + \sin x - 1$.Now,we need to evaluate $\lim_{x ightarrow 0} \frac{\Delta(x)}{x} = \lim_{x ightarrow 0} \frac{e^x + \sin x - 1}{x}$.This is a $\frac{0}{0}$ form,so we use standard limits.Using standard limits: $\lim_{x ightarrow 0} \left( \frac{e^x - 1}{x} + \frac{\sin x}{x} ight)$.$= 1 + 1 = 2$.

If $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$,then $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = $

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