Evalute: $$\lim_{x\to
0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}$$
My attempt:
I used the standard limits from the table:$$\lim_{x\to 0}\frac{\sin x}{x}=1,\;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},\;\;\lim_{x\to 0}\frac{\tan x}{x}=1,\;\;\lim_{x\to 0}\frac{e^x-1}{x}=1$$
$$$$
$L=\displaystyle\lim_{x\to 0}\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{x\arctan x}=$
$$$$
$\displaystyle\lim_{x\to 0}\left[\frac{\sin\left(e^{1-\cos^3x}-e^{1-\cos^4x}\right)}{e^{1-\cos^3x}-e^{1-\cos^4x}}\cdot\left(\frac{e^{1-\cos^3x}-1}{1-\cos^3x}\cdot\frac{1-\cos^3x}{x^2}-\frac{e^{1-\cos^4x}-1}{1-\cos^4x}\cdot\frac{1-\cos^4x}{x^2}\right)\cdot\frac{x}{\arctan x}\right]$
Substitution: $$[t=\arctan x\implies x=\tan t\;\;\&\;\; x\to 0\implies t\to 0]$$
$$\lim_{x\to 0}\frac{x}{\arctan x}\iff\lim_{t\to 0}\frac{\tan t}{t}=1$$
The next step:$$1-\cos^3x=(1-\cos x)(1+\cos x+\cos^2x)$$
$$1-\cos^4x=(1-\cos x)(1+\cos x)(1+\cos^2x)$$
Now I obtained:
$$L=1\cdot\left(1\cdot\frac{3}{2}-1\cdot 2\right)\cdot 1=-\frac{1}{2}$$
Is this correct?
Best Answer
Alternatively $$\lim_{x\to0}\dfrac{e^{1-\cos^3x}-e^{1-\cos^4x}}{x\arctan x}$$ $$=\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{\cos^4x-\cos^3x}-1}{x\arctan x}$$
$$=-\lim_{x\to0}\cos^3x\cdot\lim_{x\to0} e^{1-\cos^4x}\cdot\lim_{x\to0}\dfrac{e^{(\cos^4x-\cos^3x)}-1}{\cos^4x-\cos^3x}\cdot\lim_{x\to0}\dfrac{1-\cos x}{x\arctan x}$$
$$=-\lim_{x\to0}\dfrac{1-\cos x}{x\arctan x}$$
$$=-\lim_{x\to0}\left(\dfrac{\sin x}x\right)^2\cdot\lim_{x\to0}\dfrac x{\arctan x(1+\cos x)}$$ $$=-\dfrac1{1+1}$$