lim cosx-1/xx ->0
0
We can apply L’Hopital’s rule to evaluate this limit as we have an indeterminate form of \(\frac{0}{0}\). Taking the derivative of the numerator and denominator separately, we have:
\[\lim_{x\to 0} \frac{\cos(x)-1}{x\cdot \sin(x)} = \lim_{x\to 0} \frac{-\sin(x)}{\sin(x)+x\cdot \cos(x)}\]
Plugging in \(x=0\) directly results in the indeterminate form of \(\frac{0}{0}\), so we can apply L’Hopital’s rule again:
\[\begin{aligned}\lim_{x\to 0} \frac{-\sin(x)}{\sin(x)+x\cdot \cos(x)} &= \lim_{x\to 0} \frac{-\cos(x)}{\cos(x)-x\cdot \sin(x)} \\ &= \frac{-1}{1} = -1\end{aligned}\]
Therefore, \(\lim_{x\to 0} \frac{\cos(x)-1}{x\cdot \sin(x)} = -1\).
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