2. Find the limits:
1) $\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)$
2) $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+1}-x\right)$
3) $\lim _{n \rightarrow \infty} 2^{\prime \prime} \sin \frac{x}{2^{n}}$
4) $\lim _{x \rightarrow 1}(1-x) \operatorname{tg} \frac{\pi x}{2}$.
Solution. 1) It is not hard to see that in this example we have the difference of two infinitely large positive quantities, i.e., an indeterminate form of the type $\infty-\infty$:

$\lim _{x \rightarrow 3}\left(\frac{1}{x-3}-\frac{6}{x^{2}-9}\right)=\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-9}=\lim _{x \rightarrow 3} \frac{1}{x+3}=\frac{1}{6}$.

2) Consider two cases:

a) $\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+1}-x\right)$. We have an indeterminate form of the type $\infty-\infty$. By moving the irrationality from the numerator to the denominator, we get

$$
\begin{aligned}
\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+1}-x\right) & =\lim _{x \rightarrow+\infty} \frac{\left(\sqrt{x^{2}+1}-x\right)\left(\sqrt{x^{2}+1}+x\right)}{\sqrt{x^{2}+1}+x}= \\
& =\lim _{x \rightarrow+\infty} \frac{1}{\sqrt{x^{2}+1}+x}=0
\end{aligned}
$$

b) $\lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}+1}-x\right)=+\infty$.

3) An indeterminate form of the type $0 \cdot \infty$ can easily be reduced to an indeterminate form of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. We will reduce it to an indeterminate form of the type $\frac{0}{0}$: $\lim _{n \rightarrow \infty} 2^{n} \sin \frac{x}{2^{n}}=\lim _{n \rightarrow \infty} \frac{x \sin \frac{x}{2^{n}}}{\frac{x}{2^{n}}}=x \lim _{n \rightarrow \infty} \frac{\sin \frac{x}{2^{n}}}{\frac{x}{2^{n}}}=x \lim _{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha}=x$.
4) An indeterminate form of the type $0 \cdot \infty$ will be reduced to an indeterminate form of the type $\frac{0}{0}$:

$$
\begin{gathered}
\lim _{x \rightarrow 1}(1-x) \operatorname{tg} \frac{\pi x}{2}=\lim _{x \rightarrow 1}(1-x) \cdot \frac{\sin \frac{\pi x}{2}}{\cos \frac{\pi x}{2}}= \\
=\lim _{x \rightarrow 1} \sin \frac{\pi x}{2} \cdot \lim _{x \rightarrow 1} \frac{\frac{\pi}{2}(1-x)}{\frac{\pi}{2} \sin \left(\frac{\pi}{2}-\frac{\pi x}{2}\right)}=1 \cdot \frac{2}{\pi}=\frac{2}{\pi} .
\end{gathered}
$$

Here, we can also make a variable substitution, setting $1-x=t$, then $x=1-t$. From this, it is clear that if $x \rightarrow 1$, then $t \rightarrow 0$:

$\lim _{x \rightarrow 1}(1-x) \operatorname{tg} \frac{\pi x}{2}=\lim _{t \rightarrow 0} t \operatorname{tg}\left(\frac{\pi}{2}-\frac{\pi}{2} t\right)=\lim _{t \rightarrow 0} t \operatorname{ctg} \frac{\pi}{2} t=$

$$
=\lim _{t \rightarrow 0} \frac{t \cdot \cos \frac{\pi}{2} t}{\sin \frac{\pi}{2} t}=\lim _{t \rightarrow 0} \cos \frac{\pi}{2} t \cdot \lim _{t \rightarrow 0} \frac{\frac{\pi}{2} t}{\frac{\pi}{2} \cdot \sin \frac{\pi}{2} t}=1 \cdot \frac{2}{\pi}=\frac{2}{\pi}
$$

Find the limits: