What is the solution to this trigonometric limit problem?

AI Thread Summary
The limit problem involves finding the limit as x approaches 0 for the expression (1 - cos(x)cos(2x)cos(3x)) / (1 - cos(x)). The solution begins with rewriting the numerator, which simplifies the expression. By adding and subtracting cos(x) in the numerator, the problem becomes more manageable. This technique allows for the expansion of the terms, leading to the correct limit. The discussion concludes with the problem being solved successfully.
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Homework Statement



Find \lim_{x \to 0}\frac{1-cosxcos2xcos3x}{1-cosx}

The Attempt at a Solution



Actually my book gives this continuation \lim_{x \to 0}\frac{1-cosx+cosx[1-cos2x+cos2x(1-cos3x)]}{1-cosx} but I don't know how author arrived there. Can anyone explain it to me?

Thank you
 
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The author rewrote the numerator; if you expand all brackets, you arrive at the same expression.
 
Yeah, I realized, after I posted this thread, that author added and substracted cosx in numerator. This was the trick.

The problem is solved.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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