Question

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

Answer 1

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,
\(\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4}=\lim_{x\rightarrow0}(\frac{1}{6}+\frac{31}{310}x^2+...)\)
\(=\frac{1}{6}+\lim_{x\rightarrow0}\frac{31}{310}x^2+...\)
\(=\frac{1}{6}\)
\(\lim_{x\rightarrow0}\frac{\sec x-\cos x-x^2}{x^4}=\frac{1}{6}\)