To be calculated $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+2 x+1}+\sqrt{x^{2}+4 x+1}-\sqrt{4 x^{2}+1}\right)$.
We have already seen (for example, in problem 1300, in the 8th issue of this year¹), that if $a>0$, then the asymptotes of the hyperbola given by

$$
y=\sqrt{a x^{2}+b x+c}, \quad \text { or } \quad y^{2}-a x^{2}-b x-c=0
$$

are given by the equations: $y= \pm \sqrt{a}\left(x+\frac{b}{2 a}\right)$.

We can now rephrase this to say that the value of the expression

$$
\sqrt{a x^{2}+b x+c} \text { approaches } \sqrt{a}\left(x+\frac{b}{2 a}\right)
$$

as $x \rightarrow+\infty$, or that the limit of the difference

$$
\sqrt{a x^{2}+b x+c}-\sqrt{a}\left(x+\frac{b}{2 a}\right)
$$

is zero as $x \rightarrow+\infty$. According to this,

$$
\sqrt{x^{2}+4 x+1}=x+2+\varepsilon \quad \text { and } \quad \sqrt{4 x^{2}+1}=2 x+\varepsilon^{\prime}
$$

where $\varepsilon$ and $\varepsilon^{\prime}$ approach zero as $x \rightarrow+\infty$.

Since

$$
\sqrt{x^{2}+2 x+1}=x+1
$$

the value of the expression in question, when $x$ is sufficiently large,

$$
x+1+x+2+\varepsilon-2 x-\varepsilon^{\prime}=3+\varepsilon-\varepsilon^{\prime}
$$

If now $x \rightarrow+\infty$, then $\varepsilon \rightarrow 0$ and $\varepsilon^{\prime} \rightarrow 0$, so the limit of the expression is: 3.

Radovics György (Érseki rg. VII. o. Bp. II.)[^0]


[^0]:    ${ }^{1} 1937 / 4$. 246. p. - the editor.