I thought it might be instructuve and of interest to present an approach to evaluating the integral $\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx$ that does not rely on direct integration. To that end we proceed.
Let $f(y)$ be represented by
$$f(y)=\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx \tag1$$
Differentiating $(1)$ under the integral reveals
$$f'(y)=-\int_{-\infty}^\infty xe^{-x^2}\sin(xy)\,dx\tag2$$
Integrating by parts the integral in $(2)$ with $u=-\sin(xy)$ and $v=-\frac12e^{-x^2}$, we obtain
$$\begin{align}
f'(y)&=-\frac12y\int_{-\infty}^\infty e^{-x^2}\cos(xy)\,dx\\\\\
&=-\frac12yf(y)\tag3
\end{align}$$
From $(3)$, we see that $f(y)$ satisfies the ODE $f'(y)+\frac12yf(y)=0$, subject to $f(0)=\sqrt\pi$. The solution to this ODE is trivial and is given by
$$f(y)=\sqrt\pi e^{-y^2/4}\tag4$$
Setting $y=n$ in $(4)$ yields
$$\int_{-\infty}^\infty e^{-x^2}\cos(nx)\,dx=\sqrt\pi e^{-n^2/4}$$
Letting $n\to \infty$, we find the coveted limit is $0$.
It's a shorthand notation, but is not really "bad form" if both you and your readers understand what is really going on. What you really mean at each step where you write something like "$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} g(n)$" is
"if $\lim_{n \to \infty} g(n)$ exists, then $\lim_{n \to \infty} f(n)$ exists and has the same value". When at the end of the calculation you find that the last limit does exist,
that tells you that everything is good and you have found the limit you wanted. If you
ended up with a limit that doesn't exist, then you might not know about the original limit.
EDIT:
In particular, I note that you are using l'Hopital's rule to go from $\lim_{n \to \infty} \log\left(\frac{n-1}{n}\right)/(1/n)$ to $\lim_{n \to \infty} (1/(n^2-n))/(1/n^2)$. This is OK since the rule says if $f(n)$ and $g(n)$ both go to $0$ or $\infty$ as $n \to \infty$ and $\lim_{n \to \infty} f'(n)/g'(n) = L$ exists, then $\lim_{n \to \infty} f(n)/g(n) = L$ as well.
You shouldn't use the rule in the other direction, since it can happen that $\lim_{n \to \infty} f(n)/g(n)$ exists but $\lim_{n \to \infty} f'(n)/g'(n)$ does not.
Best Answer
Say we let
$$H(x)=\begin{cases} 0, & x < 0, \\ 1, & x > 0, \end{cases}$$
and let $H(0)$ be not defined.
Say I would like to approach $0$ on this function. However, a problem arises! Looking at the plot of the function, it is clear that if one were to approach from the right hand side, the limit is $1$, whilst if one approaches from the left, the limit is $0$ and thus the two-sided limit does not exist (both sides should be approaching the same number for this limit to exist)! This can also be easily seen by plugging in numbers:
$$H(1)=1$$ $$H(.1)=1$$ $$H(.000000000001)=1$$ etc. But, doing the same thing from the left hand side, we find
$$H(-1)=0$$ $$H(-.1)=0$$ $$H-(.000000000001)=0$$
Thus we need to define a different type of limit for functions with similar discontinuities so we may approach from either side. This limit is the "one-sided limit" and is used generally when a two-sided limit does not exist, like in the above case. $\lim_{x \to x_0^+}f(x)$ represents the right handed limit of $f(x)$ to $x_0$ whilst $\lim_{x \to x_0^-}f(x)$ represents the left hand limit. So we see that $\lim_{x \to 0} H(x)$ does not exist, but
$$\lim_{x \to 0^+}H(x)=1$$ $$\lim_{x \to 0^-}H(x)=0$$