<s>[INST] $\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ 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.
 [/INST] 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...
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