If A neq 0 and x 0,then lim _{n rightarrow | vedclass

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(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}}$.Divide the numerator and the denominator by $e^{nx}$:$

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$\frac{1}{A}$

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(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}}$.Divide the numerator and the denominator by $e^{nx}$:$= \lim _{n \rightarrow \infty} \frac{\frac{\cos x}{e^{nx}} - 1}{\frac{1}{e^{nx}} - A}$.Since $x > 0$,as $n \rightarrow \infty$,$e^{nx} \rightarrow \infty$.Thus,$\frac{\cos x}{e^{nx}} \rightarrow 0$ and $\frac{1}{e^{nx}} \rightarrow 0$.Substituting these values,we get:$= \frac{0 - 1}{0 - A} = \frac{-1}{-A} = \frac{1}{A}$.

If $A \neq 0$ and $x > 0$,then $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}} = $

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