Finding limit of a trigonometric expression

Question

Question:

Find the value of the limit of the following expression such that $x$ approaches to zero $$\frac{\cos(\sin x)-\cos x}{x^4}.$$ My attempt:

Answer 1

$$\frac{\cos(\sin x)-\cos x}{x^4}=\dfrac24\cdot\dfrac{\sin\dfrac{\sin x+x}2}{\dfrac{\sin x+x}2}\cdot\dfrac{\sin\dfrac{x-\sin x}2}{\dfrac{x-\sin x}2}\cdot\dfrac{{\sin x+x}}x\cdot\dfrac{x-\sin x}{x^3}$$

Now use $\lim_{y\to0}\dfrac{\sin y}y=1$

and from Are all limits solvable without L'Hôpital Rule or Series Expansion,

$$\lim_{t\to0}\dfrac{t-\sin t}{t^3}=\dfrac1{3!}$$

Answer 2

It's much simpler with *Taylor's expansion at order $4$.

Remember that

• $\sin x=x-\dfrac{x^3}6+o(x^4)$,
• $\cos u=1-\dfrac{u^2}2+\dfrac{u^4}{24}+o(u^4)$,

and that we can compose polynomial expansions. So in the numerator, we substitute $\;x-\dfrac{x3}6$ to $\sin x$, truncating powers of $x$ beyond degree $4$:

• $(\sin x)^2=\Bigl(x-\dfrac{x^3}6\Bigl)^2+o(x^4)=x^2-\dfrac{x^4}3+o(x^4)$,
• $(\sin x)^4=x^4+o(x^4)$,

so that finally $$\frac{\cos(\sin x)-\cos x}{x^4}=\frac{1-\dfrac{x^2}2+\dfrac{5x^4}{24}-1+\dfrac{x^2}2-\dfrac{x^4}{24}+o(x^4)}{x^4}=\frac{\dfrac{x^4}{6}+o(x^4)}{x^4}=\frac16+o(1).$$