Example 6. Calculate the limits: a) $\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)$; b) $\lim _{x \rightarrow-\infty} x^{3} e^{x} ;$ c) $\lim _{x \rightarrow \pi / 2}(\cos x)^{(x-\pi / 2)} ; \quad$ d) $\lim _{x \rightarrow 0}(1-\sin x)^{\operatorname{ctg}^{2} x}$.

A visual overview indicates the presence of different indeterminacies.
Solution.
a) $\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)=(\infty-\infty)=\lim _{x \rightarrow 1} \frac{x \cdot \ln x-x+1}{(x-1) \ln x}=$

$$
\begin{aligned}
=\left(\frac{0}{0}\right) \stackrel{!}{=} \lim _{x \rightarrow 1} \frac{\ln x+1-1}{\ln x+\frac{x-1}{x}} & =\lim _{x \rightarrow 1} x \cdot \lim _{x \rightarrow 1} \frac{1 \ln x}{x \ln x+x-1}= \\
& =1 \cdot\left(\frac{0}{0}\right) \stackrel{!}{=} \lim _{x \rightarrow 1} \frac{1 / x}{\ln x+1+1}=\frac{1}{2}
\end{aligned}
$$

Here we got rid of the factor $x$, complicating the derivative, and applied L'Hôpital's rule twice.

$$
\text { b) } \begin{aligned}
\lim _{x \rightarrow-\infty} x^{3} \cdot e^{x} & =(\infty \cdot 0)=\lim _{x \rightarrow-\infty} \frac{x^{3}}{e^{-x}}=\left(\frac{\infty}{\infty}\right) \stackrel{!}{=} \lim _{x \rightarrow-\infty} \frac{3 x^{2}}{-e^{-x}}= \\
& =\left(\frac{\infty}{\infty}\right) \stackrel{!}{=} \lim _{x \rightarrow-\infty} \frac{6 x}{e^{-x}}=\left(\frac{\infty}{\infty}\right)=\lim _{x \rightarrow-\infty} \frac{6}{-e^{-x}}=0
\end{aligned}
$$

It is recommended to compute this limit using the substitution $x=-t$ $(x \rightarrow-\infty \Rightarrow t \rightarrow+\infty)$. c) Let's find the limit of the logarithm of the expression under the limit sign. We will use the theorem on the possibility of passing to the limit under the sign of a continuous function:

$$
\begin{gathered}
\lim _{x \rightarrow \pi / 2} \ln (\cos x)^{x-\pi / 2}=\lim _{x \rightarrow \pi / 2}\left(x-\frac{\pi}{2}\right) \ln \cos x=(0 \cdot \infty)= \\
=\lim _{x \rightarrow \pi / 2} \frac{\ln \cos x}{\frac{1}{x-\frac{\pi}{2}}}=\left(\frac{\infty}{\infty}\right) \stackrel{!}{=} \lim _{x \rightarrow \pi / 2} \frac{-\sin x / \cos x}{-\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}}}= \\
=\lim _{x \rightarrow \pi / 2} \sin x \cdot \lim _{x \rightarrow \pi / 2} \frac{\left(x-\frac{\pi}{2}\right)^{2}}{\cos x}=\left(\frac{0}{0}\right) \stackrel{!}{=} \lim _{x \rightarrow \pi / 2} \frac{2\left(x-\frac{\pi}{2}\right)}{\sin x}=0
\end{gathered}
$$

Conclusion: the desired limit is $e^{0}=1$.

d) This example, like the previous one, can be solved using the basic exponential-logarithmic identity: $A=e^{\ln A}(A>0)$. Therefore, $\lim _{x \rightarrow 0}(1-\sin x)^{\operatorname{ctg}^{2} x}=\left(1^{\infty}\right)=e^{\lim _{x \rightarrow 0} \operatorname{ctg}^{2} x \cdot \ln (1-\sin x)}$. For the sake of brevity, we take the limit of the exponent separately:

$$
\begin{aligned}
\lim _{x \rightarrow 0} \cos ^{2} x \cdot \lim _{x \rightarrow 0} \frac{\ln (1-\sin x)}{\sin ^{2} x} & \stackrel{!}{=} 1 \cdot \lim _{x \rightarrow 0} \frac{-\cos x /(1-\sin x)}{2 \sin x \cos x}= \\
& =-\lim _{x \rightarrow 0} \frac{1}{2 \sin x(1-\sin x)}= \begin{cases}+\infty, & x0\end{cases}
\end{aligned}
$$

We obtained two different one-sided limits, hence the desired limit does not exist. Nevertheless, let's fix the one-sided limits:

$$
\lim _{x \rightarrow 0+0}(1-\sin x)^{\operatorname{ctg}^{2} x}=0, \quad \lim _{x \rightarrow 0-0}(1-\sin x)^{\operatorname{ctg}^{2} x}=+\infty
$$

## Exercises

1. Find the points where the tangents to the curve $y=3 x^{4}+4 x^{3}-$ $-12 x^{2}+20$ are parallel to the x-axis.

Hint. Find the points where $y^{\prime}=0$.

2. Write the equations of the tangents and normals to the curve $y=$ $=(x-1)(x-2)(x-3)$ at the points of intersection with the x-axis.
3. Write the equations of the tangent and normal to the curve $x=\frac{1+t}{t^{3}}$, $y=\frac{3}{2 t^{2}}+\frac{1}{2 t}$ at the point with rectangular coordinates $(2,2)$.

Hint, $x=2$ and $y=2$ when $t=1$.

4. Determine the intervals of monotonicity of the functions:
1) $y=x^{2}(x-3)$
2) $y=(x-3) \sqrt{x}$
3) $y=e^{x^{2}-4 x}$
4) $y=x \ln x$.
5. Investigate the functions for extrema:
1) $\left.y=x(x-1)^{2}(x-2)^{3} ; \quad 2\right) y=(x-2) \frac{8-x}{x^{2}}$;
2) $y=\frac{x^{2}-6 x+13}{x-3}$; 4) $y=x \ln ^{2} x$.

Apply L'Hôpital's rule to find the limits.
6. $\lim _{x \rightarrow 0} \frac{x-\operatorname{arctg} x}{x^{3}}$.
7. $\lim _{x \rightarrow 0} \frac{e^{x^{2}-1}}{\cos x-1}$
8. $\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}$.
9. $\lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{x}{\ln x}\right)$
10. $\lim _{x \rightarrow 0}\left(\operatorname{ctg} x-\frac{1}{x}\right) \cdot$ 11. $\lim _{x \rightarrow 0}\left(x^{2} e^{1 / x^{2}}\right)$.
12. $\lim _{x \rightarrow \pi / 2}(\operatorname{tg} x)^{2 x-\pi}$. 13. $\lim _{x \rightarrow \infty} x\left(e^{1 / x}-1\right)$. 14. $\lim _{x \rightarrow 0+0}\left(\ln \frac{1}{x}\right)^{x}$.
15. $\lim _{x \rightarrow 1-0} \ln x \cdot \ln (1-x)$.
16. $\lim _{x \rightarrow+\infty}\left(\frac{2}{\pi} \operatorname{arctg} x\right)^{x}$.
17. $\lim _{x \rightarrow 0}\left(\frac{\arcsin x}{x}\right)^{1 / x^{2}}$.
18. $\lim _{x \rightarrow 0}\left(\frac{\operatorname{arctg} x}{x}\right)^{1 / x^{2}}$. 19. $\lim _{x \rightarrow 0} \frac{\ln \sin 2 x}{\ln \sin 3 x}$.

## Answers

1. $x=0, x=1, x=-2$. 2. $y=2 x-2, y=\frac{1-x}{2} ; y=-x+2$, $y=x-2 ; y=2 x-6, y=\frac{3-x}{2}$. 3. $7 x-10 y+6=0,10 x+7 y-34=0$. 4. 1) $(0 ; 2)$, decreasing; 2$)(0 ; 1)$, decreasing; 3$)(-\infty ; 2)$, decreasing; 4) $\left(0 ; e^{-1}\right)$, decreasing. 5. 1) $y_{\min }(0,23) \approx-0,76, y_{\min }(1,4) \approx-0,05, y_{\max }(1) \approx 0$; 2) $\left.y_{\max }(3,2)=\frac{9}{16} ; 3\right) y_{\max }(1)=-4, y_{\min }(5)=4$; 4) $y_{\max }\left(e^{-2}\right)=4 e^{-2}$, $y_{\min }(1)=0.6 . \frac{1}{3} .7 .-2.8 .2 .9 .-1.10 .0 .11 . \infty ., 12.1 .13 .1 .14 .1$. 1