Let m and n be two positive integers greater | vedclass

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(B) We are given the limit: $\lim_{\alpha \rightarrow 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\frac{e}{2}$.Factor out $e$ from the numerator: $\

Vedclass · Accepted Answer

$2$

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(B) We are given the limit: $\lim_{\alpha \rightarrow 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\frac{e}{2}$.Factor out $e$ from the numerator: $\lim_{\alpha \rightarrow 0} \frac{e(e^{\cos(\alpha^n)-1} - 1)}{\alpha^m} = -\frac{e}{2}$.Using the standard limit $\lim_{x \rightarrow 0} \frac{e^x - 1}{x} = 1$,let $x = \cos(\alpha^n) - 1$. As $\alpha \rightarrow 0$,$x \rightarrow 0$.So,$\lim_{\alpha}$ ${\rightarrow 0} \frac{e(e^{\cos(\alpha^n)-1} - 1)}{\cos(\alpha^n) - 1} \cdot \frac{\cos(\alpha^n) - 1}{\alpha^m} = -\frac{e}{2}$.This simplifies to $e \cdot 1 \cdot \lim_{\alpha \rightarrow 0} \frac{\cos(\alpha^n) - 1}{\alpha^m} = -\frac{e}{2}$.Using the expansion $\cos(x) \approx 1 - \frac{x^2}{2}$,we have $\cos(\alpha^n) - 1 \approx -\frac{(\alpha^n)^2}{2} = -\frac{\alpha^{2n}}{2}$.Substituting this: $\lim_{\alpha \rightarrow 0} \frac{-\alpha^{2n}}{2\alpha^m} = -\frac{1}{2}$.For this limit to be a non-zero constant,the powers of $\alpha$ must match,so $2n = m$.Thus,$\frac{m}{n} = 2$.

Let $m$ and $n$ be two positive integers greater than $1$. If $\lim_{\alpha \rightarrow 0} \left( \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} \right) = -\left( \frac{e}{2} \right)$,then the value of $\frac{m}{n}$ is

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