$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ 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.
 A: I think the mistake is from the first step.Emm...in fact,$$\lim \frac{{\ln (1 + x\arctan x)}}{{\sqrt {1 + 2{x^4}}  - 1}} = \frac{{\ln (1 + x\arctan x)(\sqrt {1 + 2{x^4}}  + 1)}}{{2{x^4}}} = \infty , \\ \lim \frac{{{e^{{x^2}}} - 1}}{{\sqrt {1 + 2{x^4}}  - 1}} = \frac{{({e^{{x^2}}} - 1)(\sqrt {1 + 2{x^4}}  + 1)}}{{2{x^4}}} =  \infty $$
However,when $${f_1} \sim \alpha ,{f_2} \sim \beta $$ but $$\frac{\alpha }{g} = \infty or\frac{\beta }{g} = \infty $$
you can't apply
$$\frac{{{f_1} - {f_2}}}{g} \sim \frac{{\alpha  - \beta }}{g}$$
I don't know if I made it clear to you...
A: $$L=\lim_{x \to 0}\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
You need to use Taylor series:
$$L=\lim_{x \to 0}\frac{\ln (1+x (x-\dfrac {x^3}3+...)-(1+x^2+\dfrac {x^4}{2}+o(x^4))+1}{x^4}$$
$$L=\lim_{x \to 0}\frac{ (x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4))-x^2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3-\dfrac {x^4}2-\dfrac {x^4}{2}+o(x^4)}{x^4}$$
$$L=\lim_{x \to 0}\frac{ -\dfrac {x^4}3- {x^4}+o(x^4)}{x^4}$$
$$L= -\dfrac {1}3-1=-\dfrac  43$$
Where :
$$\sqrt{1+2x^4}=1+x^4+o(x^4)$$
$$e^{x^2}=1+x^2+\dfrac {x^4}2+o(x^4)$$
$$\ln (1+x)=x-\dfrac {x^2}2 +...$$
$$\arctan x = x-\dfrac {x^3}3+o(x^4)$$

for the arctan and log term
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
Remember that we have the Taylor serie for log:
$$\ln (1+z)=z-\dfrac {z^2}2$$
So that we have:
$$S=\ln (1+x (x-\dfrac {x^3}3+...)$$
$$S=\ln (1 +(x^2-\dfrac {x^4}3+...)$$
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^2-\dfrac {x^4}3+...)^2$$
We only keep terms with exponent less or equal to four...
$$S=(x^2-\dfrac {x^4}3+...)-\dfrac 12(x^4 \ \color {red}{ \text { + more terms with exponent> 4}})$$
We don't need the terms that have exponent greater than four. Because we take the limit at zero and all these terms will be zeros. They play no role. So that:
$$S=x^2-\dfrac {x^4}3-\dfrac {x^4}2+o(x^4)$$
A: 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.
