$ \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.
 A: @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it, I am able now to combine it with some unpublished computations I did.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.
A: 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
A: 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.
