The limit of the following series as X approaches \(\frac{{\rm{\p

The limit of the following series as X approaches \(\frac{{\rm{\p
| The limit of the following series as X approaches \(\frac{{\rm{\pi }}}{4}\) is \({\rm{f}}\left( {\rm{x}} \right) = {\rm{1}} - \frac{{{{\rm{x}}^2}}}{{2{\rm{\;}}!}} + \frac{{{{\rm{x}}^4}}}{{4{\rm{\;}}!}} - \frac{{{{\rm{x}}^6}}}{{6{\rm{\;}}!}} \ldots .\)

A. <span class="math-tex">\(\frac{1}{{\sqrt 2 }}\)</span>

B. 1

C. 0

D. <span class="math-tex">\(\frac{{\rm{\pi }}}{4}\)</span>

Please scroll down to see the correct answer and solution guide.

Right Answer is: A

SOLUTION

Concept: Identify the series first and then put the limits.

Calculation: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{1}} - \frac{{{{\rm{x}}^2}}}{{2{\rm{\;}}!}} + \frac{{{{\rm{x}}^4}}}{{4{\rm{\;}}!}} - \frac{{{{\rm{x}}^6}}}{{6{\rm{\;}}!}} \ldots .\)

f(x) = cos x

\({\rm{f}}{\left( {\rm{x}} \right)_{\lim {\rm{x}} \to {\rm{\;}}\frac{{\rm{\pi }}}{4}}} = {\rm{cos}}\frac{{\rm{\pi }}}{4} = \frac{1}{{\sqrt 2 }}\)