For A neq 0 and x

Question

(B) Given the limit: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.Since $x < 0$,as $n \rightarrow \infty$,$e^{n x} \rightarro

Vedclass · Accepted Answer

$\sin x$

Vedclass · Answer

(B) Given the limit: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.Since $x Dividing the numerator and denominator by $e^{n x}$ (or simply evaluating the limit as $e^{n x} \rightarrow 0$):$\lim _{n}$ ${\rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}} = \frac{\sin x - 0}{1 + A(0)} = \frac{\sin x}{1} = \sin x$.Wait,let us re-evaluate the limit expression: $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.If $x Thus,$\frac{\sin x - 0}{1 + A(0)} = \sin x$.

For $A \neq 0$ and $x < 0$,evaluate $\lim _{n \rightarrow \infty} \frac{\sin x - e^{n x}}{1 + A e^{n x}}$.

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