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Step-by-Step Solution
Step 1: Understand the problem
We are given the function
f(x) = \dfrac{\cos(\sin x) - \cos x}{x^4}
and told that it is continuous at each point in its domain. At x=0, the function is defined to be f(0) = \dfrac{1}{k}. We need to find the value of k.
Step 2: Recognize that continuity implies the limit value
For the function to be continuous at x = 0, we must have
f(0) = \lim_{x \to 0} f(x).
Hence,
\dfrac{1}{k} = \lim_{x \to 0} \dfrac{\cos(\sin x) - \cos x}{x^4}.
Step 3: Use series expansions for small x
For very small x, we use the Taylor (Maclaurin) series expansions:
\sin x = x - \dfrac{x^3}{6} + O(x^5).
\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + O(x^6).
\cos(\sin x) \approx 1 - \dfrac{(\sin x)^2}{2} + \dfrac{(\sin x)^4}{24} + \dots
Step 4: Expand \cos(\sin x) up to x^4 terms
Substitute \sin x = x - \dfrac{x^3}{6} + O(x^5) into \cos(\sin x):
\sin x = x - \dfrac{x^3}{6} + O(x^5) \quad\Rightarrow\quad (\sin x)^2 = x^2 - \dfrac{x^4}{3} + O(x^6).
Hence,
\cos(\sin x)
= 1
- \dfrac{(\sin x)^2}{2}
+ \dfrac{(\sin x)^4}{24}
+ O(x^6).
Up to x^4 terms,
(\sin x)^2 = x^2 - \dfrac{x^4}{3},
\quad
(\sin x)^4 = (x^2 - \tfrac{x^4}{3})^2 \approx x^4.
So,
\cos(\sin x)
= 1
- \dfrac{1}{2} \Bigl(x^2 - \dfrac{x^4}{3}\Bigr)
+ \dfrac{1}{24} \bigl(x^4\bigr)
+ O(x^6).
Simplifying,
\cos(\sin x)
= 1
- \dfrac{x^2}{2}
+ \dfrac{x^4}{6}
+ \dfrac{x^4}{24}
+ O(x^6)
= 1
- \dfrac{x^2}{2}
+ \dfrac{5x^4}{24}
+ O(x^6).
Step 5: Expand \cos x up to x^4 terms
\cos x
= 1
- \dfrac{x^2}{2}
+ \dfrac{x^4}{24}
+ O(x^6).
Step 6: Subtract and divide by x^4
Subtract the two expansions:
\cos(\sin x) - \cos x
= \Bigl(1 - \dfrac{x^2}{2} + \dfrac{5x^4}{24}\Bigr)
- \Bigl(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\Bigr)
+ O(x^6).
This simplifies to
\dfrac{5x^4}{24} - \dfrac{x^4}{24}
= \dfrac{4x^4}{24}
= \dfrac{x^4}{6}.
Therefore,
\lim_{x \to 0} \dfrac{\cos(\sin x) - \cos x}{x^4}
= \lim_{x \to 0} \dfrac{\tfrac{x^4}{6}}{x^4}
= \dfrac{1}{6}.
Step 7: Conclude the value of k
Since
\dfrac{1}{k}
= \lim_{x \to 0} f(x)
= \dfrac{1}{6},
we get
k = 6.
Final Answer: k = 6
