Let's calculate the following limit:

$$
\lim _{x \rightarrow \infty} x\left(\operatorname{arctg} \frac{x+1}{x+2}-\frac{\pi}{4}\right)
$$
I. solution. Let the sought limit be denoted by $H$.

$$
\text { Let } \quad u=\frac{x+1}{x+2}, \quad \text { i.e., } \quad x=\frac{2 u-1}{1-u}(x \neq-2, u \neq 1)
$$

$H$ exists if and only if $G=\lim _{u \rightarrow 1} \frac{2 u-1}{1-u}\left(\operatorname{arctg} u-\frac{\pi}{4}\right)$ exists, and in this case, clearly $H=G$ (this follows from the theorems on the limits of functions). However: $\lim _{u \rightarrow 1}(1-2 u)=-1$ and $\lim _{u \rightarrow 1} \frac{\operatorname{arctg} u-\frac{\pi}{4}}{u-1}=\lim _{u \rightarrow 1} \frac{\operatorname{arctg} u-\operatorname{arctg} 1}{u-1}$. The latter limit, however, by the definition of the derivative function:

$$
(\operatorname{arctg} u)_{u=1}^{\prime}=\left(\frac{1}{1+u_{.}^{2}}\right)_{u=1}=\frac{1}{2}
$$

Applying what we know about the limit of a product, it follows that $H=-\frac{1}{2}$.

Béla Brindza (Csongrád, Batsányi J. Gymnasium)

II. solution. It is easily seen that

$$
\operatorname{arctg} x-\operatorname{arctg} y=\operatorname{arctg} \frac{x-y}{1+x y}(x y \neq-1)
$$

and

$$
\lim _{x \rightarrow 0} \frac{\operatorname{arctg} x}{x}=1
$$

The calculation of the sought limit can then proceed as follows:

$$
H=\lim _{x \rightarrow \infty} x\left(\operatorname{arctg} \frac{x+1}{x+2}-\frac{\pi}{4}\right)=\lim _{x \rightarrow \infty} x\left(\operatorname{arctg} \frac{x+1}{x+2}-\operatorname{arctg} 1\right)
$$

By (1), this can be further transformed, and thus by (2):

$$
\begin{gathered}
H=\lim _{x \rightarrow \infty} x \operatorname{arctg} \frac{-1}{2 x+3}=\lim _{x \rightarrow \infty} x \cdot \frac{-1}{2 x+3} \cdot \frac{\operatorname{arctg} \frac{-1}{2 x+3}}{\frac{-1}{2 x+3}}= \\
=\lim _{x \rightarrow \infty} \frac{-x}{2 x+3} \cdot \frac{\operatorname{arctg} \frac{-1}{2 x+3}}{\frac{-1}{2 x+3}}=\left(\lim _{x \rightarrow \infty} \frac{-x}{2 x+3}\right)\left(\lim _{y \rightarrow 0} \frac{1}{y} \operatorname{arctg} y\right)=-\frac{1}{2}
\end{gathered}
$$

József Gubics (Székesfehérvár, Ságvári E. Vocational School)

Note. The problem can also be solved using L'Hôpital's rule (but one must ensure that the conditions for its applicability are met!)