3 years ago

Two functions, P and Q, are described as follows:

Mathematics

1 answer:

frez [133]3 years ago

7 0

I believe the answer is 4.

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Step-by-step explanation:

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zaharov [31]

Answer with Step-by-step explanation:

Given

$f(x)=\frac{1-cos(2x)}{1-cos(x)}\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-(cos^2{x}-sin^2{x})}{1-cos(x)})\\\\(\because cos(2x)=cos^2x-sin^2x)\\\\\lim_{x \rightarrow 0}f(x)=\lim_{x\rightarrow 0}(\frac{1-cos^2x}{1-cos(x)}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}(\frac{(1-cosx)(1+cosx)}{1-cosx}+\frac{sin^2x}{1-cosx})\\\\=\lim_{x\rightarrow 0}((1+cosx)+\frac{sin^2x}{1-cosx})\\\\\therefore \lim_{x \rightarrow 0}f(x)=1$

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