315. Find the limits:
1) $\lim x \operatorname{ctg} 2 x$
2) $\lim _{x \rightarrow+0} \sqrt[3]{x} \ln x$
3) $\lim (\operatorname{tg} \varphi-\sec \varphi)$; $\varphi \rightarrow \frac{\pi}{2}$
4) $\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{x}{x-1}\right)$;
5) $\lim _{t \rightarrow 0}\left(\frac{1}{\sin t}-\frac{1}{t}\right)$
Solution. By establishing that the case is $0 \cdot \infty$ or $\infty - \infty$, we transform the function into a fraction where both the numerator and the denominator simultaneously tend to zero or infinity, then apply L'Hôpital's rule:

1) $\lim _{x \rightarrow 0} x \operatorname{ctg} 2 x=\lim \frac{x}{\operatorname{tg} 2 x}=\lim \frac{1}{2 \sec ^{2} 2 x}=\frac{1}{2}$;

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-108.jpg?height=168&width=1014&top_left_y=44&top_left_x=136)
3) $\lim _{\varphi \rightarrow-\frac{\pi}{2}}(\operatorname{tg} \varphi-\sec \varphi)=\lim \frac{\sin \varphi-1}{\cos \varphi}=\lim \frac{\cos \varphi}{-\sin \varphi}=0$;
4) $\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{x}{x-1}\right)=\lim \frac{x-1-x \ln x}{(x-1) \ln x}=\lim \frac{-\ln x}{\ln x+\frac{x-1}{x}}=$

$$
:=-\lim \frac{x \ln x}{x \ln x+x-1}=-\lim \frac{1+\ln x}{2+\ln x}=\cdots \frac{1}{2}
$$

here L'Hôpital's rule is applied twice;
5) $\lim _{t \rightarrow 0}\left(-\frac{1}{\sin t}-\frac{1}{t}\right)=\lim \frac{t-\sin t}{t \sin t}=\lim \frac{1-\cos t}{\sin t+t \cos t}=$

$$
=\lim \frac{\sin t}{2 \cos t-t \sin t}=0
$$

here L'Hôpital's rule is applied twice.