Use Taylor's theorem to evaluate the following limits. lim_{xrightarrow0}frac{xsin(x)-x^2}{cos(x)-1+frac{x^2}{2}}

Question
Limits and continuity
asked 2020-10-26
Use Taylor's theorem to evaluate the following limits. \(\lim_{x\rightarrow0}\frac{x\sin(x)-x^2}{\cos(x)-1+\frac{x^2}{2}}\)

Answers (1)

2020-10-27
Evaluate limit using Taylor’s theorem.
Given:
\(\lim_{x\rightarrow0}\frac{x\sin(x)-x^2}{\cos(x)-1+\frac{x^2}{2}}\)
Taylor series expansion of trigonometric functions,
\(\sin x=\frac{x-x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\), \(-\infty
Substitute the series,
\(\lim_{x\rightarrow0}\frac{x\frac{x-x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}-1+\frac{x^2}{2}\)
\(\lim_{x\rightarrow0}\frac{(x^2-\frac{x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+...) -x^2}{\frac{1-x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)-1+\frac{x^2}{2})}\)
Cancelling the term,
\(\lim_{x\rightarrow0}\frac{\frac{-x^4}{3!}+\frac{x^6}{5!}-\frac{x^8}{7!}+...}{\frac{x^4}{4!}-\frac{x^6}{6!}+...}\)
Divide \(x^4\) in both numerator and denominator,
\(\lim_{x\rightarrow0}\frac{\frac{-1}{3!}+\frac{x^2}{5!}-\frac{x^4}{7!}+...}{\frac{1}{4!}-\frac{x^2}{6!}+..}\)
Apply limit, x tends to 0.
\(\Rightarrow\frac{\frac{-1}{3!}}{\frac{1}{4!}}\)
\(\Rightarrow\frac{-4!}{3!}\)
\(\Rightarrow\frac{4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}\)
\(\Rightarrow-4\)
0

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