Answer
The limit does not exist.
Work Step by Step
$\lim\limits_{x\to \infty} \ln(\frac{2}{x^{2}})$
We know that $\lim\limits_{x \to \infty} \frac{1}{x} = 0$.
Hence, $\lim\limits_{x \to \infty} \frac{1}{x^{2}} = 0$.
Hence, $\lim\limits_{x \to \infty} \frac{2}{x^{2}} = 0$.
Now, substitute $\frac{2}{x^{2}} = t$.
Thus, notice that when x tends to infinity, $t$ tends to 0.
Hence, this is the modified limit:
$\lim\limits_{t \to 0} \ln t$
We know that when $t = 0$, $\ln t$ does not exist ......($\ln 0$ does not exist)
But, here, $t$ is tending to 0, but not equal to 0.
Look at the graph of $\ln t$. It approaches negative infinity as $t$ approaches 0 from the right hand side (positive $t$).
However, when we say $\lim\limits_{t \to 0} \ln t$, we imply that we approach $0$ from both sides: $0+$ and $0-$. In this case, $\lim\limits_{t \to 0-} \ln t$ does not exist (look at the graph attached). Thus, in this case, the limit simply does not exist.
