what will be if we find the $\lim_{x\to \infty}(e^{-x} -1 )$ from right side such as
$\lim_{x\to \infty}(e^{-x}-1) = \lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)$
what will be if we find the $\lim_{x\to \infty}(e^{-x} -1 )$ from right side such as
$\lim_{x\to \infty}(e^{-x}-1) = \lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)$
A few comments:
It does not make sense to approach $\infty$ "from the right". What we mean by $\infty$ is a quantity which is bigger than every real number. If you draw the number line, approaching $\infty$ means we want to get as far to the right on that line as possible. The only way to approach $\infty$ is "from the left", so we instead just say $x \to \infty$, rather than $x \to \infty$ from the left.
LaTeX tip: if you use backslashes before your $lim$, you will get $\lim$. In other words, the output of $\lim$
will be $\lim$ rather than $lim$.
A few hints:
$\lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)$ is actually a double limit:
$\lim_{x\rightarrow \infty}\left( - x + \frac{x^2}2 - \frac{x^3}6 + \frac{x^4}{24}-\cdots\right)=\lim_{x\rightarrow \infty} \lim_{m \to \infty} \sum_{k=1}^m \frac{(-x)^k}{k!}$
Since in general double limits don't commute, you have to first calculate the inside limit in the RHS, and then you get the LHS..
Thus, if you calculate the RHS, you end up with the LHS.