Find the limit using Taylor-series , $f(x) =\frac{\ln(1+x^2)-x^2}{\sqrt{1+x^4}-1}$.

I calculated the limit of of the $\ln(1+x^2)$ which is equal to ${x^2} - \frac{2x^3}{3} + ...O(x^n)$

and $\ln(1+x^2) -x^2 $ = $\frac{-2x^3}{3} + ...$

but when i calculate $\sqrt{1+x^4}$ i get $0$ for first second and third derivative(when i plug in $0$) for the $4$rth derivative(after plugging in zero) i get $6$.

So Taylor series for $\sqrt{1+x^4}-1 = \frac{6x^4}{4!}+ ...$ and taking the next derivative gives again zero :( in simple i get this experession $\frac{-8x^3}{3x^4} + ..$ and $ \displaystyle{\lim_{x \to 0} f(x)}$. can someone tell me whether these steps are right or wrong?



Use the generalized binomial theorem to show that $$\sqrt{1+x^4}=(1+x^4)^{\frac 12}=1+\frac{x^4}{2}-\frac{x^8}{8}+\cdots$$

Another way could be to consider Taylor series $$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+O\left(t^3\right)$$ and replace $t$ by $x^4$.


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