Use Taylor's theorem to evaluate the following limits. limx→0xsin⁡(x)−x2cos⁡(x)−1+x22

Evaluate limit using Taylor’s theorem. Given: limx→0xsin⁡(x)−x2cos⁡(x)−1+x22 Taylor series expansion of trigonometric functions, sin⁡x=x−x33!+x55!−x77!+..., −∞ Substitute the series, limx→0xx−x33!+x55!−x77!+...1−x22!+x44!−x66!+...−1+x22 limx→0(x2−x43!+x65!−x87!+...)−x21−x22!+x44!−x66!+...)−1+x22) Cancelling the term, limx→0−x43!+x65!−x87!+...x44!−x66!+... Divide x4 in both numerator and denominator, limx→0−13!+x25!−x47!+...14!−x26!+.. Apply limit, x tends to 0. ⇒−13!14! ⇒−4!3! ⇒4⋅3⋅2⋅13⋅2⋅1 ⇒−4

"Use Taylor's theorem to evaluate the following limits...." solved and explained

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Daniaal Sanchez

Answered question

2020-10-26

Use Taylor's theorem to evaluate the following limits. $lim_{x \to 0} \frac{x \sin (x) - x^{2}}{\cos (x) - 1 + \frac{x^{2}}{2}}$

Answer & Explanation

Demi-Leigh Barrera

Skilled2020-10-27Added 97 answers

Evaluate limit using Taylor’s theorem.
Given:
$lim_{x \to 0} \frac{x \sin (x) - x^{2}}{\cos (x) - 1 + \frac{x^{2}}{2}}$
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,
$lim_{x \to 0} \frac{x \frac{x - x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . .}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + . . .} - 1 + \frac{x^{2}}{2}$
$lim_{x \to 0} \frac{(x^{2} - \frac{x^{4}}{3!} + \frac{x^{6}}{5!} - \frac{x^{8}}{7!} + . . .) - x^{2}}{\frac{1 - x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + . . .) - 1 + \frac{x^{2}}{2})}$
Cancelling the term,
$lim_{x \to 0} \frac{\frac{- x^{4}}{3!} + \frac{x^{6}}{5!} - \frac{x^{8}}{7!} + . . .}{\frac{x^{4}}{4!} - \frac{x^{6}}{6!} + . . .}$
Divide $x^{4}$ in both numerator and denominator,
$lim_{x \to 0} \frac{\frac{- 1}{3!} + \frac{x^{2}}{5!} - \frac{x^{4}}{7!} + . . .}{\frac{1}{4!} - \frac{x^{2}}{6!} + . .}$
Apply limit, x tends to 0.
$\Rightarrow \frac{\frac{- 1}{3!}}{\frac{1}{4!}}$
$\Rightarrow \frac{- 4!}{3!}$
$\Rightarrow \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1}$
$\Rightarrow - 4$

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