Taylor's series problems Can you help please me solve this by using Taylor series?
I would be grateful if you can explain how did you solve it
$$\lim_{x\to 0}{\cosh{2x\over 2+x^4}+\cos{2x\over 2+x^4}-2e^{x^4\over 2}\over \tan\sqrt{1+x^4}-\tan\sqrt{1-x^4}}$$
https://i.stack.imgur.com/xUZA1.png
 A: Let's develop every term to the fifth order:
$$\cosh \frac{2x}{2+x^4} \approx 1+ \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$\cos \frac{2x}{2 + x^4} \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{O}(x^5)$$
$$2\ e^{x^4/2} \approx 2 + x^4 + \mathcal{O}(x^5)$$
$$\sqrt{1 + x^4} \approx 1 + \frac{x^4}{2} + \mathcal{O}(x^5)$$
$$\tan\sqrt{1+x^4} \approx \tan(1)  + \frac{1}{2}\left(1 + \tan(1)^2\right)x^4 + \mathcal{O}(x^5)$$
$$\sqrt{1 - x^4} \approx 1 - \frac{x^4}{2} + \mathcal{O}(x^5)$$
$$\tan\sqrt{1+x^4} \approx \tan(1)  +\left(- \frac{1}{2} - \frac{\tan(1)^2}{2}\right)x^4 + \mathcal{O}(x^5)$$
Accodino to that, we rewrite the function as
$$\lim_{x\to 0} \frac{1+ \frac{x^2}{2} + \frac{x^4}{24} + 1 - \frac{x^2}{2} + \frac{x^4}{24} - 2 - x^4}{\tan(1)  + \frac{1}{2}\left(1 + \tan(1)^2\right)x^4 -\tan(1)  -\left(- \frac{1}{2} - \frac{\tan(1)^2}{2}\right)x^4 }$$
$$\lim_{x\to 0} \frac{-\frac{22}{24}}{\frac{2 + 2\tan(1)^2}{2}}$$
$$\lim_{x\to 0} \frac{-\frac{11}{12}}{\tan(1)^2 + 1}$$
Now, generally speaking
$$\tan^2(x) + 1 = \frac{\sin^2(x)}{\cos^2(x)} + 1 = \frac{\sin^2(x) + \cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}$$
In your case
$$1 + \tan^2(1) = \frac{1}{\cos^2(1)}$$
hence the final result is
$$\boxed{-\frac{11}{12}\cos^2(1)}$$
