<s>[INST] $ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not able.
 [/INST] L'Hospital once the fraction becomes
$$\frac{2\sin^2{x}+\tan^2{x}-3x^2}{7x^6}$$
Still an indeterminate
L'hospital 6 more times and rearranging the fraction becomes
$$\frac{4\cos{2x}}{315}+\sec^8{x}-\frac{4\sec^6{x}}{3}+\frac{2\sec^4{x}}{5}-\frac{4\sec^2{x}}{315}$$
And the limit for $x\to 0$ is $1-\frac{4}{3}+\frac{2}{5}=\frac{1}{15}$
7 times l'Hospital is so much of a pain: Taylor in this case is more convenient
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