$ \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.

@Paramanand. Thank you very much. Your method is powerfull. After reading
it, I am able now to combine it with some unpublished computations I did.
My favorite method is to do not use LHR nor Taylor expansion whenever
possible and to came back all the computations to the basic limits only. For
example, to compute the limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}}
\end{equation*}
one can make use the following basic limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}=\frac{1}{3},\ \ \ \ \ \ \text{and
}\ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6},
\end{equation*}
as follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}} &=&\lim_{x\rightarrow 0}%
\frac{(\tan x-x)+(x-\sin x)}{x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{%
x-\sin x}{x^{3}} \\
&=&\frac{1}{3}+\frac{1}{6}=\frac{1}{2}.
\end{eqnarray*}
To compute the limit $A=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\tan x+2\sin
x-3x}{x^{5}}$ I suggest the same way but to push further one more order. By
LHR one can use the following basic limits:
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}=\frac{2}{%
15},\ \ \ \ \ \ and\ \ \ \ \ \ \ \lim_{x\rightarrow 0^{+}}\frac{\sin x-x+%
\frac{1}{6}x^{3}}{x^{5}}=\frac{1}{120}.
\end{equation*}
Hence, 
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{\tan x+2\sin x-3x}{x^{5}} &=&\lim_{x%
\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})+2(\sin x-x+\frac{1}{6}%
x^{3})}{x^{5}} \\
&=&\lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}%
+2\lim_{x\rightarrow 0^{+}}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} \\
&=&\frac{2}{15}+2\left( \frac{1}{120}\right)  \\
&=&\frac{3}{20}.
\end{eqnarray*}
Now I can say that taking into account the development of Paramanand Singh,
and the remark above, it is (finally) possible to compute the original
limit by making use of only the basic following limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\tan x-x-\frac{1}{3}x^{3}}{x^{5}} &=&\frac{2}{15}
\\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}%
.
\end{eqnarray*}
The computations could be conducted as follows
\begin{eqnarray*}
\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{\sin x-x}{x^{3}}\right)
\left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left( 
\frac{\sin x\tan x-x\sin x+\sin ^{2}x-x^{2}}{x^{6}}\right)  \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( 
\frac{\tan x-x}{x^{3}}\right)  \\
&&+\left( \frac{x\tan x+\sin ^{2}x-2x^{2}}{x^{6}}\right)  \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( 
\frac{\tan x-x}{x^{3}}\right)  \\
&&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x+2\sin x-3x}{%
x^{5}}\right)  \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left( 
\frac{\tan x-x}{x^{3}}\right)  \\
&&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x-x-\frac{1}{3%
}x^{3}}{x^{5}}\right) +2\left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}%
\right) 
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{-1%
}{6}\right) \left( \frac{1}{3}\right) \left( \frac{1}{1}\right) +\left( 
\frac{-1}{6}\right) \left( \frac{1}{3}\right) +\left( \frac{-1}{6}\right)
\left( \frac{-1}{6}\right) +\left( \frac{2}{15}\right) +2\left( \frac{1}{120}%
\right)  \\
&=&\frac{1}{15}.
\end{eqnarray*}
It is like decomposing a non-prime number into a product of prime
numbers...Here, the basic limits are the prime numbers, and performing
'a' decomposition is a state of art.
Thanks again to Paramanand.
