3.2.1. Find the limits of functions:

a) $\lim _{x \rightarrow \pm \infty} \sqrt{x^{2}-2 x-1}-\sqrt{x^{2}-7 x+3}$ (indeterminate form $\infty-\infty$ ).
Solution. Multiply and divide the given expression by $\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}$:

$$
\begin{aligned}
& \sqrt{x^{2}-2 x-1}-\sqrt{x^{2}-7 x+3} \quad \sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3} \\
& \lim _{x \rightarrow \pm \infty} \sqrt{\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}}= \\
& =\lim _{x \rightarrow \pm \infty} \frac{x^{2}-2 x-1-x^{2}+7 x-3}{\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}}=\lim _{x \rightarrow \pm \infty} \frac{5 x-4}{\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}}=
\end{aligned}
$$

(Indeterminate form $\frac{ \pm \infty}{\infty}$; divide the numerator and denominator by $x$ and place the sign " $\pm$ " before $\lim$):

$$
= \pm \lim _{x \rightarrow \pm \infty} \frac{5-\frac{4}{x}}{\sqrt{1-\frac{2}{x}-\frac{1}{x^{2}}}+\sqrt{1-\frac{7}{x}+\frac{3}{x^{2}}}}= \pm \frac{5}{\sqrt{1}+\sqrt{1}}= \pm \frac{5}{2}
$$

b)

$$
\lim _{x \rightarrow 2} \frac{8 x^{2}-20 x+8}{\sqrt{4 x}-\sqrt{2 x+4}}=\frac{8 \cdot 2^{2}-20 \cdot 2+8}{\sqrt{4 \cdot 2}-\sqrt{2 \cdot 2+4}}=\frac{0}{0}-\text { indeterminate form }
$$

Factor the polynomial in the numerator and multiply the numerator and denominator by $(\sqrt{4 x}+\sqrt{2 x}+4)$:

$$
\begin{aligned}
& \left.\lim _{x \rightarrow 2} \frac{8 x^{2}-20 x+8}{\sqrt{4 x}-\sqrt{2 x+4}}=\lim _{x \rightarrow 2} \frac{8(x-2) x-\frac{1}{2}(\sqrt{4 x}+\sqrt{2 x+4})}{(\sqrt{4 x}-\sqrt{2 x+4}) \cdot(\sqrt{4 x}+\sqrt{2 x+4}}\right)= \\
& =8 \lim _{x \rightarrow 2} \frac{(x-2) x-\frac{1}{2}}{4 x-2 x-4} \cdot \lim _{x \rightarrow 2}(\sqrt{4 x}+\sqrt{2 x+4})= \\
& =8 \lim _{x \rightarrow 2} \frac{(x-2) x-\frac{1}{2}}{2(x-2)} \cdot(\sqrt{4 \cdot 2}+\sqrt{2 \cdot 2+4})=4 \cdot 2 \sqrt{8} \lim _{x \rightarrow 2} x-\frac{1}{2}= \\
& =16 \sqrt{2} \cdot 2-\frac{1}{2}=24 \sqrt{2}
\end{aligned}
$$

c)

$$
\lim _{x \rightarrow 0} \frac{\cos 2 x-\cos 6 x}{1-\cos 4 x}=\frac{\cos 0-\cos 0}{1-\cos 0}=\frac{1-1}{1-1}=\frac{0}{0}-\text { indeterminate form }=
$$

(Apply the sum-to-product formula in the numerator and the half-angle formula in the denominator)

$$
=\lim _{x \rightarrow 0} \frac{-2 \sin 4 x \cdot \sin (-2 x)}{2 \sin ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{\sin 4 x \cdot \sin 2 x}{\sin ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 2 x}=\lim _{x \rightarrow 0} \frac{4 x}{2 x}=\frac{4}{2}=2
$$

Replace the numerator and denominator with equivalent infinitesimals: $\sin 4 x \sim 4 x ; \sin 2 x \sim 2 x$;
d) $\lim _{x \rightarrow \infty} \frac{2 x-1}{2 x+1}^{2 x}=\lim _{x \rightarrow \infty} \frac{(2 x+1)-2}{2 x+1}^{2 x}=\lim _{x \rightarrow \infty} 1-\frac{2}{2 x+1}^{2 x} \Rightarrow$

Let $\alpha=-\frac{2}{2 x+1}$, then $2 x=-\frac{2}{\alpha}-1$, and as $x \rightarrow \infty$, we have $\alpha \rightarrow 0$.

Then

$$
\begin{gathered}
\Rightarrow \lim _{\alpha \rightarrow 0}(1+\alpha)^{-\frac{2}{\alpha}-1}=\lim _{\alpha \rightarrow 0}(1+\alpha)^{-\frac{2}{\alpha}} \cdot \lim _{\alpha \rightarrow 0}(1+\alpha)^{-1}= \\
=\lim _{\alpha \rightarrow 0}(1+\alpha)^{\frac{1}{\alpha}} \quad \cdot(1+0)^{-1}=(e)^{-2} \cdot 1=e^{-2}
\end{gathered}
$$

e)

$$
C=\lim _{x \rightarrow \pi}(\cos 2 x)^{\operatorname{ctg}^{2} 4 x}=(\cos 2 \pi)^{\operatorname{ctg}^{2} 4 \pi}=1^{\infty} \text { — indeterminate form }
$$

Let $\cos 2 x=1+\alpha(x)$, where $\lim _{x \rightarrow \pi} \alpha(x)=0$ and thus,

$$
\begin{gathered}
C=\lim _{x \rightarrow \pi}(1+\alpha(x))^{\frac{1}{\alpha(x)}}{ }^{\alpha(x) \cdot \operatorname{ctg}^{2} 4 x}=\lim _{x \rightarrow \pi} e^{\alpha(x) \cdot \operatorname{ctg}^{2} 4 x}= \\
=e^{\lim _{x \rightarrow \pi} \alpha(x) \cdot \operatorname{ctg}^{2} 4 x}=e^{\lim _{x \rightarrow \pi}(\cos 2 x-1) \cdot \operatorname{ctg}^{2} 4 x}
\end{gathered}
$$

Calculate:

$$
\begin{aligned}
A=\lim _{x \rightarrow \pi}(\cos 2 x-1) \cdot \operatorname{ctg}^{2} 4 x=\lim _{x \rightarrow \pi}\left(-2 \sin ^{2} x\right) \cdot \frac{\cos ^{2} 4 x}{\sin ^{2} 4 x}= \\
=-2 \lim _{x \rightarrow \pi} \cos ^{2} 4 x \cdot \lim _{x \rightarrow \pi} \frac{\sin ^{2} x}{\sin ^{2} 4 x}=-2 \cos ^{2} 4 \pi \cdot \lim _{x \rightarrow \pi} \frac{\sin x}{\sin 4 x}
\end{aligned}
$$

Let $t=x-\pi$, then $x=\pi+t$ and if $x \rightarrow \pi$, then $t \rightarrow 0$. Then

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-132.jpg?height=110&width=707&top_left_y=1522&top_left_x=379)

$$
\begin{aligned}
& =-2 \lim _{t \rightarrow 0} \frac{\sin t}{\sin 4 t}^{2}=-2 \lim _{t \rightarrow 0} \frac{\sin t}{t} \frac{4 t}{\sin 4 t} \cdot \frac{1}{4}^{2}= \\
& =-2 \cdot \frac{1}{16} \lim _{t \rightarrow 0} \frac{\sin t}{t} \cdot \lim _{\beta \rightarrow 0} \frac{\beta}{\sin \beta}{ }^{2}=-\frac{1}{8}(1 \cdot 1)^{2}=-\frac{1}{8}
\end{aligned}
$$

Let $\beta=4 t$ and if $t \rightarrow 0$, then $\beta \rightarrow 4 \cdot 0 \rightarrow 0$. We have $C=e^{A}=e^{-\frac{1}{8}}$ finally $\lim _{x \rightarrow \pi}(\cos 2 x)^{\operatorname{ctg}^{2} 4 x}=e^{-\frac{1}{8}}$.