Finding the limit without L'Hospital's rule. Find the value of $n$ for the given limit.
$$\lim\limits_{x \to 1}\frac{π/4-\tan^{-1}x}{e^{\sin(\ln x)}-x^n} =\frac 18.$$
ATTEMPT:
I tried expanding the denominator by Maclaurin's series, but the term $e^{\sin(\ln x)}$ has a very complicated expansion. Is there any other way to solve this without L'Hospital's rule?
 A: This is a very good question (+1 for OP for the same) as it requires us to know how fast the numerator and denominator of the given expression tend to $0$ as $x \to 1$. To simplify things we need to put $x = 1 + h$ and let $h \to 0$. Next consider the numerator $$\frac{\pi}{4} - \tan^{-1}(1 + h) = \tan^{-1}(1) - \tan^{-1}(1 + h)$$ which clearly looks like $\phi(a) - \phi(a + h)$ and is expected to be almost like $-h\phi'(a)$ where $\phi(x) = \tan^{-1}(x), a = 1, \phi'(a) = 1/2$. It is better to divide the numerator and denominator by $h$ (had $\phi'(a) = 0$ we would have to divide by a higher power of $h$ but here just dividing by $h$ is OK) so that our expression becomes $$F(h) = \dfrac{\dfrac{\pi}{4} - \tan^{-1}(1 + h)}{\exp(\sin(\log(1 + h))) - (1 + h)^{n}} = \dfrac{\dfrac{\tan^{-1}(1) - \tan^{-1}(1 + h)}{h}}{\dfrac{\exp(\sin(\log(1 + h))) - (1 + h)^{n}}{h}} = \frac{f(h)}{g(h)}$$ where we can see that
\begin{align}
\lim_{h \to 0}f(h) &= \lim_{h \to 0}\frac{\tan^{-1}(1) - \tan^{-1}(1 + h)}{h}\notag\\
&= \lim_{h \to 0}\dfrac{\tan^{-1}\left(\dfrac{1 - (1 + h)}{1 + 1\cdot(1 + h)}\right)}{h}\notag\\
&= -\lim_{h \to 0}\dfrac{\tan^{-1}\left(\dfrac{h}{2 + h}\right)}{\dfrac{h}{2 + h}}\cdot\frac{1}{2 + h}\notag\\
&= -1\cdot 1\cdot\frac{1}{2} = -\frac{1}{2}
\end{align}
Since $F(h) = f(h)/g(h) \to 1/8$ as $h \to 0$ it follows that $$\lim_{h \to 0}g(h) = -\frac{1}{2}\cdot 8 = -4$$ i.e. $$\lim_{h \to 0}\frac{\exp(\sin(\log(1 + h))) - (1 + h)^{n}}{h} = -4$$ We can do some simplification for calculation of limit of $g(h)$ as follows
\begin{align}
-4 &= \lim_{h \to 0}g(h)\notag\\
&= \lim_{h \to 0}\frac{\exp\{\sin(\log(1 + h))\} - (1 + h)^{n}}{h}\notag\\
&= \lim_{h \to 0}\frac{\exp\{\sin(\log(1 + h))\} - \exp\{n\log(1 + h)\}}{h}\notag\\
&= \lim_{h \to 0}\exp\{n\log(1 + h)\}\cdot\frac{\exp\{\sin(\log(1 + h)) - n\log(1 + h)\} - 1}{h}\notag\\
&= \lim_{h \to 0}1\cdot\frac{\exp\{\sin(\log(1 + h)) - n\log(1 + h)\} - 1}{\sin(\log(1 + h)) - n\log(1 + h)}\cdot\dfrac{\sin(\log(1 + h)) - n\log(1 + h)}{h}\notag\\
&= \lim_{h \to 0}1\cdot\dfrac{\sin(\log(1 + h)) - n\log(1 + h)}{h}\notag\\
&= \lim_{h \to 0}\dfrac{\sin(\log(1 + h))}{h} - n\cdot\dfrac{\log(1 + h)}{h}\notag\\
&= \lim_{h \to 0}\dfrac{\sin(\log(1 + h))}{\log(1 + h)}\cdot\frac{\log(1 + h)}{h} - n\cdot 1\notag\\
&= 1\cdot 1 - n = 1 - n
\end{align} Clearly we now have $n = 5$. As can be seen this solution avoids the use of L'Hospital's Rule and Taylor series and uses only the standard limits and rules of limits.
A: For $x=1+h$, we have $$\ln x=h-\frac12 h^2+O(h^3),$$
so $$\sin\ln x=h-\frac12 h^2+O(h^3)$$
and $$\exp\sin\ln x = 1+(h-\frac12 h^2)+\frac12(h-\frac12 h^2)^2+O(h^3)=1+h+O(h^3) $$ 
and finally in the denominator
$$ \exp\sin\ln x-x^n=(1-n)h-{n\choose2}h^2+O(h^3).$$
In the numerator
$$\arctan(1+h)=\frac\pi4+\frac12 h+O(h^2). $$ 
Thus for $n\ge2$ we have
$$\frac{ \frac\pi4-\arctan x}{\exp\sin\ln x-x^n}=\frac{-\frac12h+O(h^2)}{(1-n)h+O(h^2)}=\frac1{2(n-1)}+O(h) $$
(and for $n=1$ there is divergence). Apparently we want $n=5$. (And apparently $O(h^3)$ in the denominator was overkill)
