Find $\lim_{x\to +\infty}\frac{\sqrt {2x^2+3}}{4x+2}$ and $\lim_{x\to -\infty}\frac{\sqrt {2x^2+3}}{4x+2}$. I was reading about limits of functions from Problems of Calculus in One Variable written by I.A Maron. There was an example given in the book:
Find $\lim_{x\to +\infty}\frac{\sqrt {2x^2+3}}{4x+2}$ and $\lim_{x\to -\infty}\frac{\sqrt {2x^2+3}}{4x+2}$.
The solution given in the book is as follows:

However, I dont get how they conclude "From this it follows, incidentally, that $\lim_{x\to \infty}\frac{\sqrt {2x^2+3}}{4x+2}$ does not exist." ? I am not quite getting it...
For a more accurate reference of the question, I am attaching a picture of how the question was presented in the book:

$$
\text { Noting that } \sqrt{x^2}=\left\{\begin{array}{cl}
x & \text { if } x \geq 0 \\
-x & \text { if } x<0
\end{array}\right.
$$
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =\lim _{x \rightarrow+\infty} \frac{\frac{1}{\sqrt{x^2}} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =\lim _{x \rightarrow+\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =\frac{\sqrt{2}}{4} \\
& =\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{x \rightarrow -\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =-\lim _{x \rightarrow-\infty} \frac{-\frac{1}{x} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =-\lim _{x \rightarrow-\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =-\frac{\sqrt{2}}{4} \\
& =-\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
