# Calculus - Evalute limit without using L'Hôpital rule

I tried to solve the following limit by the L'Hôpital Rule: $$\lim_{x \to \pi/4} \frac{{\tan(x)^{\cot(x)} - \cot(x)^{\tan(x)}}}{x-\pi/4}$$

Taking $$L$$ to be the said limit, by L'Hôpital Rule,

$$L = \lim_{x \to \pi/4} \tan(x)^{\cot(x)}\ln\tan{x}(-\csc^2x) + {\tan(x)^{\cot(x) - 1}}\cot{x}\sec^2x$$

$$- \cot(x)^{\tan(x)} \ln(\cot x) \sec^2{x} + \cot(x)^{\tan(x)-1}\tan x \csc^2 x$$

$$\Rightarrow L = 4$$

Now, this method is quite involved, as the differentiation of the functions is not very simple. This compels me to think that there could be a better method to solve this question. Any help in that pursuit is appreciated.

• It is $\frac{0}{0}$ form, use L'Hospital rule, see here Commented Nov 7, 2023 at 7:18
• You can also substitute $x-\frac{\pi}{4}=h$ . So, $x=\frac{\pi}{4}+h$ . Now substitute this value of $x$ in the above limit. Commented Nov 8, 2023 at 13:46

First, by change of variable $$y=\tan x$$, your limit becomes $$L=\lim_{y\to1} \frac{y^{1/y} - (1/y)^y}{(\arctan y)-\pi/4}.$$ The denominator is asymptotically equivalent to $$(y-1)\arctan'(1)=(y-1)/2$$.
Letting $$h:=y-1\to0$$, the numerator is \begin{align}e^{(1+h)^{-1}\ln(1+h)}-e^{-(1+h)\ln(1+h)}&=e^{(1+o(1))(h+o(h))}-e^{(1+o(1))(-h+o(h))}\\&=e^{h+o(h)}-e^{-h+o(h)}\\ &\sim 2h.\end{align} The answer is therefore $$L=\lim_{h\to0}\frac{2h}{h/2}=4.$$