<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] Let the desired limit be denoted by $L$.
We have via LHR $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}} = \frac{1}{6}\tag{1}$$ and $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} = \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}} = \frac{1}{3}\tag{2}$$ and we also have $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{3}$$ Multiplying the 3 limits above we get
\begin{align}
&\lim_{x \to 0}\frac{(x - \sin x)(\tan x - x)\sin x}{x^{7}} = \frac{1}{18}\notag\\ &\Rightarrow\lim_{x \to 0}\frac{(x\sin x\tan x - x^{2}\sin x - \sin^{2}x\tan x + x\sin^{2}x)}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x - x^{3} + x^{3} - \sin^{2}x\tan x + x\sin^{2}x}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{x\sin x\tan x - x^{2}\sin x + x\sin^{2}x - x^{3}}{x^{7}} - \frac{\sin^{2}x\tan x - x^{3}}{x^{7}} = \frac{1}{18}\notag\\
&\Rightarrow\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} - L = \frac{1}{18}\notag\\
\end{align}
Our job is done if we can show that $$\lim_{x \to 0}\frac{\sin x\tan x - x\sin x + \sin^{2}x - x^{2}}{x^{6}} = \frac{11}{90}\tag{4}$$ Multiplying $(1)$ and $(2)$ we get $$\lim_{x \to 0}\frac{x\tan x - x^{2} - \sin x\tan x + x\sin x}{x^{6}}= \frac{1}{18}\tag{5}$$ Adding $(4)$ and $(5)$ we see that our proof is complete if we show that $$\lim_{x \to 0}\frac{x\tan x + \sin^{2}x - 2x^{2}}{x^{6}} = \frac{8}{45}\tag{6}$$ Squaring $(1)$ we get $$\lim_{x \to 0}\frac{\sin^{2}x + x^{2} - 2x\sin x}{x^{6}} = \frac{1}{36}\tag{7}$$ Subtracting $(7)$ from $(6)$ we see that proof is complete if we show that $$\lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}= \frac{3}{20}\tag{8}$$ It is this limit which we will calculate using LHR as follows
\begin{align}
A &= \lim_{x \to 0}\frac{\tan x + 2\sin x - 3x}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sec^{2} x + 2\cos x - 3}{5x^{4}}\text{ (apply LHR)}\notag\\
&= \lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{5x^{4}\cos^{2}x}\notag\\
&= \frac{1}{5}\lim_{x \to 0}\frac{1 + 2\cos^{3} x - 3\cos^{2}x}{x^{4}}\notag\\
&= \frac{1}{5}\lim_{x \to 0}\frac{(\cos x - 1)^{2}(2\cos x + 1)}{x^{4}}\notag\\
&= \frac{3}{5}\lim_{x \to 0}\left(\frac{1 - \cos x}{x^{2}}\right)^{2}\notag\\
&= \frac{3}{5}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{3}{20}
\end{align}
Thus the proof is complete by application of LHR three times (once in proof of $(1)$, $(2)$, $(8)$ each). Also note that if you know the result $(1)$ then result $(2)$ can be derived from $(1)$ by subtraction and noting that the limit $$\lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}$$ can be calculated without LHR very easily. See this question. So in reality we only need two application of LHR for this problem.
Update: While dealing with limit expressions of type $\lim_{x \to 0}f(x)/x^{n}$ for large $n$ (here $n = 7$), I have often found it useful to multiply several well knows limits of type $g(x)/x^{m}$ with smaller values of $m$ to get something like $h(x)/x^{n}$. Expectation is that some terms of $f(x)$ match with those of $h(x)$ and a subtraction would cancel these terms. Also it is expected that resulting expression will be simplified to $p(x)/x^{r}$ when $r < n$. Continue this till we get very small values of exponent of $x$ in denominator. Here for example I have reduced an expression with $x^{7}$ to finally an expression with $x^{5}$ in denominator. See this technique applied to $$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ here. Another application of the same technique can be found here as well.
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