Can I accept my student's solution of $\lim _{x \to \infty} \ln x-2x$ 

Can I accept my student's solution of $$\lim_{ x \to \infty} \ln x-2x$$


I gave the above limit to my student. This is the way she has done.
$$\begin{aligned}
L= & \lim _{x \rightarrow \infty} \ln x-2 x=\lim _{x \rightarrow \infty} x\left(\frac{\ln x}{x}-2\right) \\
& \lim _{x \rightarrow \infty} \frac{\ln x}{x}-2=\lim _{x \rightarrow \infty} \frac{1}{x}-2=-2 \: \text{(By L'Hopital's Rule)}\\
\Rightarrow \quad & L=\lim _{x \rightarrow \infty} x \times-2=-\infty
\end{aligned}$$
But I suggested this way:
$$\begin{gathered}
x=\frac{1}{t} \\
\Rightarrow L=\lim _{t \rightarrow 0^{+}}-\ln t-\frac{2}{t} \\
\Rightarrow \quad L=-\lim _{t \rightarrow 0^{+}} \frac{2+t \ln t}{t} \\
\lim _{t \rightarrow 0^{+}} t \ln t=0 \: \text{(Again by L'Hopital's Rule)}\\
\Rightarrow L=-\infty
\end{gathered}$$
 A: If we are being really pedantic, I think both solutions have a small "gap." Either solution wants to use a result of calculus which says something like the following:
(M) If $a_n\to a$ and $b_n\to b$, then $a_nb_n\to ab$.
(S) If $a_n\to a$ and $b_n\to b$, then $a_n + b_n\to a + b$.
Your student would like to use (M) because they know $\ln x/x -2 \to -2$, and $x\to \infty$, so by (M), we should have the limit $-\infty$. Your solution would like to use (S) because you know $t\ln t\to 0$, and $2$ is constant, so by (S), $2 + t\ln t\to 2$. (I presume this is what you had in mind.)
The thing is, the formal laws (M) and (S) work fine when $a$ and $b$ are both finite real numbers. When either of $a$ or $b$ is $\pm\infty$, things get more complicated, and so I would say both solutions need some more explanation as to how they get the final answer. If anything, I think your student's solution is clearer because it's easier to justify that (M) holds when exactly one of the the limits $a$ or $b$ is finite and nonzero, and the other is $+\infty$.
The problem I have with your solution is that you know $t\ln t \to 0$, but you are also dividing by $t$, so overall we are still in a situation where the fully written out solution needs something like an appeal to L'Hospital's rule, or an estimate of the size of $\ln t + 2/t$ as $t\to 0^+$.
A: Rewrite:
$$  \lim_{x \to \infty}\left(- 2 x + \ln{\left(x \right)}\right)  = \color{red}{\lim_{x \to \infty} \ln{\left(x e^{- 2 x} \right)}} 
$$
Move the limit under the logarithm:
$$\color{black}{\lim_{x \to \infty} \ln{\left(x e^{- 2 x} \right)}} = \color{black}{\ln{\left(\lim_{x \to \infty} x e^{- 2 x} \right)}}$$
After:
$$\ln{\left(\color{black}{\lim_{x \to \infty} x e^{- 2 x}} \right)} = \ln{\left(\lim_{x \to \infty} \frac{x}{e^{2 x}} \right)}$$
Since we have an indeterminate form of type $∞/∞$
, we can apply the l'Hopital's rule observing that
$$\ln{\left(\color{black}{\lim_{x \to \infty} \frac{\frac{d}{dx}\left(x\right)}{\frac{d}{dx}\left(e^{2 x}\right)}} \right)} = \ln{\left(\color{black}{\lim_{x \to \infty} \frac{e^{- 2 x}}{2}} \right)}\longrightarrow \ln (\to0^+)\longrightarrow -\infty$$
Hence:
$$\lim_{x \to \infty}\left(- 2 x + \ln{\left(x \right)}\right)=-\infty$$
