I'm struggling with the following limit $$\lim_{x \to \infty} e^x\big(1 + \sin(x)\big).$$
First of all I checked the solution using Wolfram. To my surprise Wolfram says that this limit exists and is equal to $\infty$. This goes against my intuition being honest.
I decided to try to show that this limit does not exist using Heine definition. I found the following sequences $$x_n = n\pi, \qquad y_n = \bigg(2n + \frac{3}{2}\bigg)\pi.$$
Of course both $x_n$ and $y_n$ tend to $\infty$ whenever $n \to \infty$. Moreover we have $\sin(x_n) = 0$ and $\sin(y_n) = -1$ for all $n \in \mathbb{N}$.
The first limit is easy, $$\lim_{n \to \infty} e^{x_n}\big(1 + \sin(x_n)\big) = \infty.$$
The problems appears when computing the next limit, which is $$\lim_{n \to \infty} e^{y_n}\big(1 + \sin(y_n)\big).$$ We clearly have an indeterminate form, i.e. $\infty \times 0$. How can I solve my problem? Does this limit exist?
The given limit does NOT exist (and Wolfram is wrong).
As you already noted, $x_n = n\pi\to +\infty$ and $$\lim_{n \to \infty} e^{x_n}\big(1 + \sin(x_n)\big) =+\infty.$$ On the other hand, if $y_n = \bigg(2n + \frac{3}{2}\bigg)\pi\to +\infty$ then, for any integer $n$, $$e^{y_n}\big(1 + \sin(y_n)\big)=e^{(2n + \frac{3}{2})\pi}\big(1 -1\big)=0$$ and therefore (no indeterminate form here!) $$\lim_{n \to \infty} e^{y_n}\big(1 + \sin(y_n)\big)=0.$$ So, along two sequences which go to $+\infty$, we obtain two different limits of $e^x\big(1 + \sin(x)\big)$, therefore $\lim_{x \to \infty} e^x\big(1 + \sin(x)\big)$ does not exist.