Find the limit $\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4(x+1)}$

Question

I need to find $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}$. I tried using the following: \begin{align*} \ln(1+x)&\approx x,\\ \sin(x)&\approx x-\frac{x^3}{2},\\ \cos(x)&\approx 1-\frac{x^2}{2!}+\frac{x^4}{4!}, \end{align*} I managed to get $\displaystyle \lim _{x\to 0}\frac{\cos x-1+\dfrac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}=\lim _{x\to 0}\frac{\left(\dfrac{\left(-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}\right)}{x^{4}}+\dfrac{x^{2}-\frac{x^{4}}{3!}}{2x^{4}}\right)}{\left(\dfrac{\ln\left(1+x\right)-x}{x}+1\right)^{4}}$ but I can't see how to use that $\displaystyle \lim _{x\to 0}\frac{\ln\left(1+x\right)-x}{x}=0$ in here.

Am I missing something? I didn't learn the little-$\mathcal{O}$ notation yet.

Thanks

Answer 1

Write: $Q(x) =\dfrac{\cos x -1 +\dfrac{x}{2}\cdot\sin x}{\ln^4(1+x)} = \dfrac{-2\sin^2\left(\frac{x}{2}\right)+2\cdot\left(\dfrac{x}{2}\right)^2\cdot\dfrac{\sin x}{x}}{\ln^4(1+x)} = \dfrac{\dfrac{x^2}{2}}{\ln^2(1+x)}\cdot\dfrac{\left(\dfrac{\sin x}{x}-\dfrac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2}\right)}{\ln^2(1+x)}= f(x)g(x)$. Observe that when $x \to 0 \implies f(x) \to \dfrac{1}{2}$, and for $g(x)$, to make it simpler put $x=2y \implies g(x) = h(y) = \dfrac{\dfrac{\sin(2y)}{2y} - \dfrac{\sin^2y}{y^2}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \dfrac{\cos y - \dfrac{\sin y}{y}}{\ln^2(1+2y)}=\dfrac{\sin y}{y}\cdot \left(\dfrac{2y}{\ln(1+2y)}\right)^2\cdot \dfrac{1}{4}\cdot \dfrac{y\cos y - \sin y}{y^3}\to 1\cdot 1\cdot \dfrac{1}{4}\cdot \dfrac{-1}{3}= \dfrac{-1}{12}\implies f(x)g(x) \to \dfrac{-1}{24} \implies \displaystyle \lim_{x \to 0} Q(x) = \dfrac{-1}{24}$.

Answer 2

You only need to add to your equalities the $\log$ $$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \textrm{h.o.t}\\ x \sin x = x^2 - \frac{x^4}{6} + \textrm{h.o.t}\\ \log(1+x) = x + \textrm{h.o.t, so} \\ \log^4(1+x) = x^4 + \textrm{h.o.t}$$

Now calculate carefully and see that $$\frac{\cos x-1 - \frac{x}{2} \sin x}{\log^4(1+x)} =\ldots = \frac{-\frac{x^4}{24} + \textrm{h.o.t}}{x^4 + \textrm{h.o.t}}$$ Now divide numerator and denominator by $x^4$, and get your result. You were on the right track!

Answer 3

You can use $$ \lim_{x\to0}\frac{\ln(1+x)}{x}=1 $$ to handle the limit. In fact \begin{eqnarray} &&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{\ln ^4\left(x+1\right)}\\ &=&\lim _{x\to 0}\frac{\cos x-1+\frac{x}{2}\cdot \sin x}{x^4}\bigg[\frac{x}{\ln \left(x+1\right)}\bigg]^4\\ &=&\lim _{x\to 0}\frac{-\sin x+\frac{1}{2} \sin x+\frac{x}{2}\cos x}{4x^3}\\ &=&\lim _{x\to 0}\frac{-\frac{1}{2} \cos x+\frac{1}{2}\cos x-\frac{x}{2}\sin x}{12x^2}\\ &=&\lim _{x\to 0}\frac{-\sin x}{24x}\\ &=&-\frac{1}{24}. \end{eqnarray}