limit of the complex sequences How can we find $\lim _{n\to \infty }\left(\frac{\sqrt[n]{in}}{2^n+in!}\right)$?
Can we use the fact $\lim_{n\rightarrow \infty }\left(z_n\right)=z_0\leftrightarrow \lim _{n\to \infty }\left(\left|z_n-z_0\right|\right)=0$ ?
 A: First one question: how do you define $\sqrt[n]{in}$?
Providing you answer this question, you'll be able to prove that:
$$\lim\limits_{n \to \infty} \vert \sqrt[n]{in} \vert = 1 \text{ and } \lim\limits_{n \to \infty} \vert 2^n+in! \vert = +\infty$$ hence your limit exists and is equal to zero.
A: HINT:
$$\lim_{n\to\infty}\space\frac{\sqrt[n]{in}}{2^n+in!}=$$
$$\lim_{n\to\infty}\space\frac{\left(in\right)^{\frac{1}{n}}}{2^n+in!}=$$
$$\lim_{n\to\infty}\space\exp\left(\ln\left(\frac{\left(in\right)^{\frac{1}{n}}}{2^n+in!}\right)\right)=$$
$$\lim_{n\to\infty}\space\exp\left(\frac{\ln(in)}{n}-\ln(2^n+in!)\right)=$$
$$\exp\left(\lim_{n\to\infty}\space\frac{\ln(in)}{n}-\lim_{n\to\infty}\space\ln(2^n+in!)\right)=$$
$$\exp\left(\lim_{n\to\infty}\space\frac{\frac{\text{d}}{\text{d}n}\left(\ln(in)\right)}{\frac{\text{d}}{\text{d}n}\left(n\right)}-\lim_{n\to\infty}\space\ln(2^n+in!)\right)=$$
$$\exp\left(\lim_{n\to\infty}\space\frac{\frac{1}{n}}{1}-\lim_{n\to\infty}\space\ln(2^n+in!)\right)=$$
$$\exp\left(\lim_{n\to\infty}\space\frac{1}{n}-\lim_{n\to\infty}\space\ln(2^n+in!)\right)=$$
$$\exp\left(0-\lim_{n\to\infty}\space\ln(2^n+in!)\right)=$$
$$\exp\left(-\lim_{n\to\infty}\space\ln(2^n+in!)\right)$$
