121. Find $\lim _{x \rightarrow 2} \frac{x^{2}+x+2}{x^{2}+2 x+8}$.
Solution. Let's check if the denominator of the fraction does not turn into zero when $x=2$: we have $2^{2}+2 \cdot 2+8=16 \neq 0$. Substituting the limit value of the argument, we find

$$
\lim _{x \rightarrow 2} \frac{x^{2}+x+2}{x^{2}+2 x+8}=\frac{2^{2}+2+2}{2^{2}+2 \cdot 2+8}=\frac{8}{16}=\frac{1}{2}
$$

Now let's consider examples where the application of limit properties becomes possible only after some preliminary transformations.