algebra precalculus - 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?

Answer

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)