1. Depending on the non-negative parameter $k$, determine the limit:

$$
\lim _{x \rightarrow 0}\left(\frac{\tan k x}{x}\right)^{\frac{1}{x^{2}}}
$$
1. First solution. First of all, we have

$$
\lim _{x \rightarrow 0} \frac{\operatorname{tg} k x}{x}=\lim _{x \rightarrow 0} \frac{\frac{\sin k x}{\cos k x}}{x}=\lim _{x \rightarrow 0} \frac{\sin k x}{k x} \cdot \lim _{x \rightarrow 0} \frac{k}{\cos k x}=1 \cdot k=k
$$

Further, we have $\lim _{x \rightarrow 0} \frac{1}{x^{2}}=\infty$, so we get that for $k>1$ it holds $\lim _{x \rightarrow 0}\left(\frac{\operatorname{tg} k x}{x}\right)^{\frac{1}{x^{2}}}=$ $\infty$, and for $0 \leqslant k < 1$ it holds $\lim _{x \rightarrow 0}\left(\frac{\operatorname{tg} k x}{x}\right)^{\frac{1}{x^{2}}}=0$. Therefore, the final result is:

$$
\lim _{x \rightarrow 0}\left(\frac{\operatorname{tg} k x}{x}\right)^{\frac{1}{x^{2}}}=\begin{cases}
\infty, & k>1 \\
1, & k=1 \\
0, & 0 \leqslant k < 1
\end{cases}
$$

Second solution. We only consider the case $k=1$ (the rest follows as in the previous solution). We can write

$$
\lim _{x \rightarrow 0}\left(\frac{\operatorname{tg} x}{x}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0}\left(e^{\ln \frac{\operatorname{tg} x}{x}}\right)^{\frac{1}{x^{2}}}=\lim _{x \rightarrow 0} e^{\frac{1}{x^{2}} \ln \frac{\operatorname{tg} x}{x}}=e^{\lim _{x \rightarrow 0} \frac{1}{x^{2}} \ln \frac{\operatorname{tg} x}{x}}
$$

so it is sufficient to compute $\lim _{x \rightarrow 0} \frac{1}{x^{2}} \ln \frac{\operatorname{tg} x}{x}$. We proceed as follows (applying L'Hôpital's rule wherever indicated by $\stackrel{\text { „̈ }}{=}$):

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\ln \frac{\operatorname{tg} x}{x}}{x^{2}} \stackrel{n n}{=} \lim _{x \rightarrow 0} \frac{\frac{x}{\operatorname{tg} x} \cdot \frac{\frac{1}{\cos ^{2} x} x-\operatorname{tg} x}{x^{2}}}{2 x}=\lim _{x \rightarrow 0} \frac{\frac{x}{\cos ^{2} x}-\operatorname{tg} x}{2 x^{2} \operatorname{tg} x}=\lim _{x \rightarrow 0} \frac{\frac{x}{\cos ^{2} x}-\frac{\sin x}{\cos x}}{2 x^{2} \frac{\sin x}{\cos x}}=\lim _{x \rightarrow 0} \frac{x-\sin x \cos x}{2 x^{2} \sin x \cos x}=\operatorname{li}_{x-}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_841f52b4ffb037ad799bg-31.jpg?height=101&width=1214&top_left_y=1033&top_left_x=364)

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\sin 2 x}{\sin 2 x+4 x \cos 2 x-2 x^{2} \sin 2 x} \stackrel{\text { N. }}{=} \lim _{x \rightarrow 0} \frac{2 \cos 2 x}{2 \cos 2 x+4 \cos 2 x-8 x \sin 2 x-4 x \sin 2 x-4 x}
\end{aligned}
$$

thus we obtain the same result as in the previous solution.