6.61. $\lim _{x \rightarrow+0}\left(\frac{\sin x}{x}\right)^{1 / x^{4}}$.

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6.61. $\lim _{x \rightarrow+0}\left(\frac{\sin x}{x}\right)^{1 / x^{4}}$.
Solution.

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## Let's find separately

$$
\begin{aligned}
& \lim _{x \rightarrow+0} \frac{(\ln \sin x-\ln x)^{\prime}}{\left(x^{4}\right)^{\prime}}=\lim _{x \rightarrow+0} \frac{\operatorname{ctg} x-\frac{1}{x}}{4 x^{3}}=\lim _{x \rightarrow+0} \frac{(x \operatorname{ctg} x-1)^{\prime}}{\left(4 x^{4}\right)^{\prime}}= \\
& =\lim _{x \rightarrow+0} \frac{\operatorname{ctg} x-\frac{x}{\sin ^{2} x}}{16 x^{3}}=\lim _{x \rightarrow+0} \frac{(\cos x \sin x-x)^{\prime}}{\left(16 x^{3} \sin ^{2} x\right)^{\prime}}= \\
& =\lim _{x \rightarrow+0} \frac{(1 / 2 \sin 2 x-x)^{\prime}}{\left(16 x^{3} \sin ^{2} x\right)^{\prime}}=\lim _{x \rightarrow+0} \frac{\cos 2 x-1}{48 x^{2} \sin ^{2} x+32 x^{3} \sin x \cos x}= \\
& =\lim _{x \rightarrow+0} \frac{2 \sin ^{2} x}{48 x^{2} \sin ^{2} x+32 x^{3} \sin x \cos x}= \\
& =\lim _{x \rightarrow+0} \frac{(\sin x)^{\prime}}{\left(24 x^{2} \sin x+16 x^{3} \cos x\right)^{\prime}}= \\
& =\lim _{x \rightarrow+0} \frac{\cos x}{48 x \sin x+24 x^{2} \sin x+48 x^{2} \cos x-16 x^{3} \sin x}=\infty .
\end{aligned}
$$

$$
\text { Hence } \lim _{x \rightarrow+0}\left(\frac{\sin x}{x}\right)^{1 / x^{4}}=\infty \text {. }
$$

## INDETERMINACY OF THE FORM " $\infty^{0}$ "