Find limlimits_(n→∞)\[e^(nπ i)-(-1)^n]n+dfrac in\

Find $\ \lim\limits_{n\to\infty}\left\{\left[e^{n\pi i}-(-1)^n\right]n+\dfrac in\right\}$.

Step 1 · Simplify $e^{n\pi i}$

$$ e^{n\pi i}=(e^{i\pi})^n=(-1)^n $$

using Euler's identity $e^{i\pi}=-1$.

Step 2 · The bracket vanishes identically

$$ \left[e^{n\pi i}-(-1)^n\right]n=\left[(-1)^n-(-1)^n\right]n=0 $$

for every integer $n$ — not just in the limit.

Step 3 · What remains

$$ \left[e^{n\pi i}-(-1)^n\right]n+\frac in=0+\frac in=\frac in $$

Step 4 · Take the limit

$$ \boxed{\lim_{n\to\infty}\frac in=i\cdot\lim_{n\to\infty}\frac1n=i\cdot0=0} $$

Verification

$\left|\dfrac in\right|=\dfrac1n\to0$, so the sequence's modulus vanishes, confirming convergence to $0$.