The integral value of n for which lim _{x rig | vedclass

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(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$.Using Taylor series expansions near $x=0$:$\cos x = 1 - \f

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$3$

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(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$.Using Taylor series expansions near $x=0$:$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$e^x = 1 + x + \frac{x^2}{2!} + \dots$Substituting these into the expression:$(\cos x - 1) = -\frac{x^2}{2} + O(x^4)$$(\cos x - e^x) = (1 - \frac{x^2}{2} + O(x^4)) - (1 + x + \frac{x^2}{2} + O(x^3)) = -x - x^2 + O(x^3)$Thus,the numerator is $(-\frac{x^2}{2} + O(x^4))(-x - x^2 + O(x^3)) = \frac{x^3}{2} + O(x^4)$.For the limit to be a finite non-zero real number,the power of $x$ in the denominator must match the lowest power of $x$ in the numerator.Therefore,$n = 3$.

The integral value of $n$ for which $\lim _{x \rightarrow 0} \frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is a finite non-zero real number is

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