Use Taylor series to evaluate the following limits. lim_{xrightarrow0}frac{sec x-cos x-x^2}{x^4} (Hint: text{The Maclaurin series for sec x is }1+frac{x^2}{2}+frac{5x^4}{24}+frac{61x^6}{720}+...)

Question
Limits and continuity
Use Taylor series to evaluate the following limits.
$$\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4} \ (Hint: \text{The Maclaurin series for sec x is }1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}+...)$$

2021-02-01
Given that:
$$\text{Let f(x)}=\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4}$$
Taylor's series,
$$f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!}(x-c)^2+...+\frac{f^n(c)}{n!}(x-c)^n+R_n(x)$$
Where
$$R_n(x)=\frac{f^{(n+1)}}{(n+1)!}(x-c)^{n+1}$$
To find the Taylor's series,
The Maclaurin series for sec x is
$$\sec x=1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}+...$$
The Maclaurin series for sec x is
$$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+...$$
$$\sec x-\cos x=(1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}+...)-(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+...)$$
$$=x^2+\frac{1}{6}x^4+\frac{31}{310}x^6+...$$
$$\sec x-\cos x-x^2=\frac{1}{6}x^4+\frac{31}{310}x^6+...$$
Then,
$$\frac{\sec x-\cos x-x^2}{x^4}=\frac{1}{6}+\frac{31}{310}x^2+...$$
Then,
$$f(x)=\frac{1}{6}+\frac{31}{310}x^2+...$$
Then,
To get the Taylor series is $$\frac{1}{6}+\frac{31}{310}x^2+...$$
To find limit as x approaches to 0
$$\frac{\sec x-\cos x-x^2}{x^4}=\frac{1}{6}+\frac{31}{310}x^2+...$$
Taking limit on both side as $$x\rightarrow0$$
To get,
$$\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4}=\lim_{x\rightarrow0}(\frac{1}{6}+\frac{31}{310}x^2+...)$$
$$=\frac{1}{6}+\lim_{x\rightarrow0}\frac{31}{310}x^2+...$$
$$=\frac{1}{6}$$
$$\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4}=\frac{1}{6}$$

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