1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

Hint

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 $x^4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.