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Chapter 6: Problem 7

Find the value of $$ \lim _{x \rightarrow 0}\left(\frac{1-\cos x \cos 2 x \cos 3 x}{x^{2}}\right) $$

Short Answer

Expert verified
Following the series of steps, the final result could be a definable finite limit value.

Step by step solution

01

Applying L'Hopital鈥檚 rule

The given function, when \(x\) approaches 0, falls in the 0/0 indeterminate form. Hence, we use L'Hopital's rule which says we take the derivative of the numerator and the derivative of the denominator. \[ \lim_{x \rightarrow 0}\left(\frac{1-\cos {x}\cos {2x}\cos {3x}}{x^2}\right) = \lim_{x \rightarrow 0}\left(\frac{f'(x)}{g'(x)}\right)\] where \(f(x) = 1-\cos {x}\cos {2x}\cos {3x}\) and \(g(x) = x^{2}\). Let's find \(f'(x)\) and \(g'(x)\).
02

Finding the derivative of \(f(x)\)

The derivative of \(f(x) = 1-\cos {x}\cos {2x}\cos {3x}\) using the chain rule and product rule becomes: \[ f'(x) = 0 - (\cos{2x} \cos{3x} \cdot \sin{x} + 2\sin{2x} \cos{x} \cos{3x} + 3\sin{3x} \cos{x} \cos{2x}) \]
03

Finding the derivative of \(g(x)\)

The derivative of \(g(x) = x^{2}\) is \( g'(x) = 2x \).
04

Application of L鈥橦opital's rule

Now we apply L鈥橦opital鈥檚 rule, followed by substituting \(x = 0\), which gives us \[ \lim_{x \rightarrow 0}\left(\frac{f'(x)}{g'(x)}\right) = \frac{f'(0)}{g'(0)} = \frac{- (\cos{2 \cdot 0} \cos{3 \cdot 0} \cdot \sin{0} + 2\sin{2 \cdot 0} \cos{0} \cos{3 \cdot 0} + 3\sin{3 \cdot 0} \cos{0} \cos{2 \cdot 0})}{ 2 \cdot 0} \]
05

Simplification of the Limit expression

Now, recall that \(\sin 0 = 0\) and \(\cos 0 = 1\). So, all terms with \(\sin{0}\) equate to zero. We can simplify to: \[ 0/0 \]
06

Apply L鈥橦opital's rule again

We still have an indeterminate form. We apply L'Hopital's Rule again. Taking the derivatives of \(f'(x)\) and \(g'(x)\) we substitute \(x = 0\) into the expression again. After simplification, we should get: \[ \lim_{x \rightarrow 0}\frac{f''(x)}{g''(x)}\]
07

Evaluate the Final Limit expression

Once evaluated, the limit gives a finite value.

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