95. Find $\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}$.
Solution. To apply the property of the limit of a quotient, we need to check whether the limit of the denominator is not zero when $x=4$. Since $\lim _{x \rightarrow 4}(x-3)=$[^3]$=\lim _{x \rightarrow 4} x-\lim 3=4-3=1 \neq 0$, in this case, we can use property 4:

$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\frac{\lim _{x \rightarrow 1}\left(x^{2}-2 x\right)}{\lim _{x \rightarrow 4}(x-3)}
$$

But $\lim _{x \rightarrow 4}\left(x^{2}-2 x\right)=\left(\lim _{x \rightarrow 4} x\right)^{2}-2 \lim _{x \rightarrow 4} x$ and, therefore,

$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\frac{\left(\lim _{x \rightarrow 4} x\right)^{2}-2 \lim _{x \rightarrow 4} x}{\lim _{x \rightarrow 4} x-\lim 3}=\frac{4^{2}-2 \cdot 4}{4-3}=8
$$

96-98. Find the limits: