Evaluate $ \lim_{ x\to 0} \frac{ \left(\arctan(2x)-2\arctan( x)\right)^x}{\mathrm e^{x^2}-1-\sin^2(x)}$

Question

It is obvious that we use Taylor’s expansion. But what shall I do with the power of $x$ in numerator? In denominator I expand $\sin^2(x)$ as multiplying two expanded $\sin$?

Answer 1

As pointed in other answer the numerator makes sense only when $x\to 0^{-}$. Putting $x=-t$ we can see that the numerator can be expressed as $$\left(\arctan\left(\frac{2t^{3}}{1+3t^{2}}\right)\right)^{-t} $$ Next we will make use of the following standard limits $$\lim_{x\to 0}\frac{\arctan x} {x} =1,\lim_{x\to 0^{+}}x^{x}=1$$ Let $z=2t^3/(1+3t^{2})$ and then $z\to 0^{+}$ and we can write the numerator as $$\left(\frac{\arctan z} {z} \right) ^{-t}\cdot \frac{(1+3t^{2})^{t}}{2^{t}(t^t)^{3}}$$ Via the above mentioned standard limits we can see that the numerator tends to $1$ as $t\to 0^{+}$. The denominator clearly tends to $0$ and remains positive therefore the desired limit is $\infty$. Most textbooks would say that the limit does not exist or the expression diverges to $\infty $.

Answer 2

HINT Use $[f(x)]^{g(x)}=\mathrm{e}^{g(x)\log f(x)}$

Answer 3

Hint for the numerator part :

$$(\arctan2x-2\arctan x)^x = e^{x\ln(\arctan2x-2\arctan x)}$$

Answer 4

Note that since

$$\arctan 2x-2\arctan x=2x-\frac{8x^3}{3}-2x+\frac{2x^3}{3}+o(x^3)=-2x^3+o(x^3)$$

the limit is well defined only for $x\to0^-$.

To handle with positive quantities set $x=-y$ with $y\to0^+$.

Thus:

$$\frac{\left(\arctan 2x -2\arctan x\right)^x}{\mathrm e^{x^2}-1-\sin^2x}=\frac{\left( \arctan(-2y)-2\arctan(-y)\right)^{-y}}{\mathrm e^{y^2}-1-\sin^2 (-y)}=\frac{\left( 2\arctan y-\arctan 2y\right)^{-y}}{\mathrm e^{y^2}-1-\sin^2 y}$$

$$\left( 2\arctan y-\arctan 2y\right)^{-y}=\left(2y^3+o(y^3)\right)^{-y}=e^{-y\log(2y^3+o(y^3))}\to e^0=1$$

and:

$$e^{y^2}-1-\sin^2 y=1+y^2+\frac{y^4}{2}+o(y^4)-1-\left(y-\frac{y^3}{6}+o(y^3)\right)^2=y^2+\frac{y^4}{2}+o(y^4)-y^2+\frac{y^4}{3}+o(y^4)=\frac{5y^4}{6}+o(y^4)\to 0$$

Finally:

$$\frac{\left(\arctan 2x -2\arctan x\right)^x}{\mathrm e^{x^2}-1-\sin^2x}\to +\infty$$