$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$
But the final result should be $-\frac{4}{3}$.
Any help would be appreciated.

The handling of denominator is easy. Multiplying the expression under limit with $$\frac{\sqrt{1+2x^4}+1}{\sqrt{1+2x^4}+1}$$ we can see that the limit in question is equal to the limit of the expression $$\frac{\log(1+x\arctan x) - e^{x^2}+1}{x^4}$$ Split this expression into three parts as $$\frac{\log(1+x\arctan x) - x\arctan x} {(x\arctan x) ^2}\cdot\frac{(\arctan x) ^2}{x^2}+\frac{\arctan x - x}{x^3}-\frac{e^{x^2}-1-x^2}{x^4}$$ For the first term substitute $u=x\arctan x$ and conclude that it tends to $-1/2$. You can also find the limit of second term easily as $-1/3$. For third term just put $t=x^2$ and observe that it tends to $1/2$. The desired answer is thus $$-\frac{1}{2}-\frac{1}{3}-\frac{1}{2}=-\frac{4}{3}$$

The issue with your approach is the wrong use of equivalents ($\sim$ notation). Unless you are familiar with the rules of this notation avoid using it and instead stick to the usual limit laws/theorems/rules.
