The integer n for which mathop {lim }limits_{ | vedclass

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{(\cos x - 1)(\cos x - e^x)}{x^n}$.Using Taylor series expansions near $x = 0$:$\cos x = 1 - \frac{

Vedclass · Accepted Answer

$3$

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(C) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{(\cos x - 1)(\cos x - e^x)}{x^n}$.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} + \dots) - (1 + x + \frac{x^2}{2} + \dots) = -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 number,the power of $x$ in the denominator must match the lowest power of $x$ in the numerator.Therefore,$n = 3$.

The integer $n$ for which $\mathop {\lim }\limits_{x \to 0} \,\frac{(\cos x - 1)(\cos x - e^x)}{x^n}$ is a finite non-zero number is

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