Find: $\lim_{x\to0}\frac{\ln(1+x^2)-x^2+\frac{x^4}{2}}{\sin x^6}$ How do I even approach such a question? Obviously l'Hospital rule would be a mess and I don't know how to even start. Most of the limits we deal with are way simpler than this.
 A: L'Hopital:
$$\lim\limits_{x\to 0}\frac{-2x+2x^3+\frac{2x}{x^2+1}}{6x^5\cos(x^6)}=\lim\limits_{x\to 0}\frac{\frac{2x^5}{x^2+1}}{6x^5\cos(x^6)}=\lim\limits_{x\to 0}\frac{2x^5}{(x^2+1)(6x^5\cos(x^6))}\\=\lim\limits_{x\to 0}\frac{\overbrace{\sec(x^6)}^{\to 1}}{3\underbrace{(x^2+1)}_{\to 1}}=\frac 1 3$$
A: How about using Series expansion, $$\ln(1+z)=-\sum_{r=1}^\infty\dfrac{(-z)^r}r$$
and $\lim_{h\to0}\dfrac{\sin h}h=1$
A: Try this
$$\ln(1+x^2) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} + \mathcal{O}(x^7)$$
and
$$\sin x^6 = x^6 + \mathcal{O}(x^7)$$
A: Set $y=x^2$ and ignore the sine (thanks to the limit of the ratio $\sin(t)/t$).
$$\lim_{x\to0}\frac{\ln(1+x^2)-x^2+\frac{x^4}{2}}{\sin(x^6)}=\lim_{y\to0}\frac{\ln(1+y)-y+\frac{y^2}{2}}{\sin(y^3)}=\lim_{y\to0}\frac{\ln(1+y)-y+\frac{y^2}{2}}{y^3}.$$
If you know the Taylor's development of $\ln(1+y)=y-\frac{y^2}2+\frac{y^3}3-\frac{y^4}4\cdots$, you recognize the first two terms which are compensated, and the answer is the coefficient of the cubic one,
$$\frac13.$$
Otherwise, L'Hospital comes to the rescue,
$$\lim_{y\to0}\frac{\frac1{1+y}-1+y}{3y^2}=\lim_{y\to0}\frac{y^2}{3y^2(1+y)}.$$
No mess.
