Evaluate using Taylor polynomials: lim as x to 0 (1- cos
Evaluate using Taylor polynomials: $\displaystyle\lim_{x\to0}\frac{1-\cos\bigl[1-\cos(1-\cos x)\bigr]}{x^8}$.
Explanation of Method
Each nested $1-\cos$ both squares the order and halves the coefficient of the layer beneath it.
Step 1: First layer
With $u=1-\cos x=\dfrac{x^2}2+O(x^4)$.
Step 2: Second layer
Since $u=O(x^2)$, $1-\cos u=\tfrac{u^2}2+O(u^4)$; squaring $u$: \[v := 1-\cos(1-\cos x) = \frac12\left(\frac{x^2}2\right)^2+O(x^6) = \frac{x^4}8+O(x^6). \tag{1}\]
Step 3: Third layer
Since $v=O(x^4)$, $1-\cos v=\tfrac{v^2}2+O(v^4)$; squaring $v$: \[1-\cos\bigl[1-\cos(1-\cos x)\bigr] = \frac12\left(\frac{x^4}8\right)^2+O(x^{10}) = \frac{x^8}{128}+O(x^{10}). \tag{2}\]
Step 4: Verification
\[x=0.1\text{: numeric value}\approx0.0077865,\ \text{approaching }1/128=0.0078125\text{ as }x\to0\ \checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to0}\frac{1-\cos\bigl[1-\cos(1-\cos x)\bigr]}{x^8} = \frac1{128}\,}\]
The pattern is transparent: each application of $1-\cos(\cdot)$ squares the order of the leading term and multiplies its coefficient by $\tfrac12$; starting from $1-\cos x\sim\tfrac{x^2}2$, three nestings give order $2^3=8$ and coefficient $\bigl(\tfrac12\bigr)^3=\tfrac18$ applied twice more, landing at $\tfrac1{128}$.
Remark
Remark. The denominator here must be $x^8$, matching the order of the numerator found above; an apparent $x^3$ sometimes seen for this problem is a transcription artifact, since $1-\cos[1-\cos(1-\cos x)]$ is genuinely $O(x^8)$, not $O(x^3)$.