\(\lim_{{x \to 0}} \limits\) \(\frac{cos(sin x) - cos x }{x^4}\) is equal to :
\(\lim_{{x \to 0}} \limits\) \(\frac{cos(sin x) - cos x }{x^4}\) is equal to :
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{6}\)
\(\frac{1}{12}\)
The Correct Option is C
Solution and Explanation
The correct answer is (C) : \(\frac{-11}{9}\)
\(\lim_{{x \to 0}} \limits\) \(\frac{cos(sinx) - cosx}{x^4}\) = \(\lim_{{x \to 0}} \limits\) \(\frac{2sin(x + sinx).sin(\frac{x - sinx}{2})}{x^4}\)
= \(\lim_{{x \to 0}} \limits\) \(2.\frac{(\frac{( x + sinx }{2})(\frac{x-sinx}{2})}{x^4}\)
= \(\lim_{{x \to 0}} \limits\) \(\frac{1}{2}.\) \((\frac{(x+x -\frac{x^3}{3!}+\frac{x^5}{5!}....)(x-x+\frac{x^3}{3!}....}{x^4})\)
= \(\lim_{{x \to 0}} \limits\) \(\frac{1}{2}.\) \((2-\frac{x^2}{3!}+\frac{x^4}{5!}....)(\frac{1}{3!}-\frac{x^2}{5!}-1)\)
= \(\frac{1}{6}\)
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Concepts Used:
Limits of Trigonometric Functions
Assume a is any number in the general domain of the corresponding trigonometric function, then we can explain the following limits.
We know that the graphs of the functions y = sin x and y = cos x detain distinct values between -1 and 1 as represented in the above figure. Thus, the function is swinging between the values, so it will be impossible for us to obtain the limit of y = sin x and y = cos x as x tends to ±∞. Hence, the limits of all six trigonometric functions when x tends to ±∞ are tabulated below:
