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:
 A: This use of $\infty$ is actually very common for complex analysis. And if we consider the limit at $\infty$ to be over the complex numbers, the author is completely correct - the fact that the real limits at $\pm\infty$ are different shows that the complex limit at $\infty$ cannot exist. That the author appears to be treating the limit at $\infty$ as a real limit is unusual, but not unheard of. Once you've learned the complex equivalent, it is obvious that the same concept can be applied to the real numbers to have them curl around and approach a single $\infty$ at both ends.
In $\Bbb R^3$, consider the plane $z = 0$ to be the complex plane, with $(x,y, 0)\equiv x + iy$. Now consider the unit sphere $S^2 = \{(x,y,z)\mid x^2+y^2+z^2=1\}$, and label $(0,0,1)$ as $n$, the "north pole". For each complex number $w$, the line passing through $n$ and $w$ will intersect the sphere in one other point $w'$. If $|w| = 1$, then $w$ and $w'$ coincide. And every point the sphere other than $n$ corresonds to a unique complex number in this way. You can view the points on the sphere as being the complex numbers in this fashion. But we have one extra point $n$ on the sphere. Points in neighborhoods of $n$ are those with large absolute value, without regard to direction, with the absolute values getting larger as the neighborhood of $n$ gets smaller. Thus it makes sense to consider the north pole to be a single infinite complex number that one arrives at no matter how they grow without bound. It is labeled just "$\infty$". The sphere consisting of $\Bbb C \cup \{\infty\}$ is called "the Riemann sphere" after its inventor and is familiar to every student of complex analysis.
Applying the same concept to the unit circle in the plane, with the real numbers acting at the $x$-axis, wraps the real numbers into a circle with a single $\infty$ as the pole.
And for this expression, since the limit at $-\infty$ is not the same as the limit at $+\infty$, it approaches different values on each side of $\infty$ and that limit cannot exist.
A: $$
\text { Noting that } \sqrt{x^2}=\left\{\begin{array}{cl}
x & \text { if } x \geq 0 \\
-x & \text { if } x<0
\end{array}\right.
$$
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =\lim _{x \rightarrow+\infty} \frac{\frac{1}{\sqrt{x^2}} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =\lim _{x \rightarrow+\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =\frac{\sqrt{2}}{4} \\
& =\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
$$
\begin{aligned}
\lim _{x \rightarrow -\infty} \frac{\sqrt{2 x^2+3}}{4 x+2} & =-\lim _{x \rightarrow-\infty} \frac{-\frac{1}{x} \sqrt{2 x^2+3}}{4+\frac{2}{x}} \\
& =-\lim _{x \rightarrow-\infty} \frac{\sqrt{2+\frac{3}{x^2}}}{4+\frac{2}{x}} \\
& =-\frac{\sqrt{2}}{4} \\
& =-\frac{1}{2 \sqrt{2}}
\end{aligned}
$$
