114. Calculate $\lim _{x \rightarrow 1} \frac{2 x^{3}+4 x-3}{x+4}$.
Solution. When $x=1$, the fraction $\frac{2 x^{3}+4 x-3}{x+4}$ is defined, since its denominator is not zero. Therefore, to compute the limit, it is sufficient to substitute the argument with its limiting value. Then we get

$$
\lim _{x \rightarrow 1} \frac{2 x^{3}+4 x-3}{x+4}=\frac{2 \cdot 1^{3}+4 \cdot 1-3}{1+4}=\frac{3}{5}
$$

The specified rule for calculating limits cannot be applied in the following cases:

1) if the function is undefined at $x=a$;
2) if the denominator of the fraction becomes zero when substituting $x=a$;
3) if both the numerator and the denominator of the fraction become zero or infinity when substituting $x=a$.

In such cases, limits of functions are found using various artificial techniques.