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I have to find if the following two functions are or not uniformly continuous:

$a) \ f:[0,1]\rightarrow R, f(x)=\frac{1}{x^2-x-2}$

$b) \ f:R\rightarrow R,f(x) = \sin(x) - cos(x)$

My way of thinking for each of them goes as follows:

a) $f$ is continuous on [0,1] and [0,1] is a compact set, thus $f$ is uniformly continuous

b) $f$ id derivable and $f'(x)=\cos(x) - \sin(x)$; however both $sin$ and $cos$ lie somewhere in [-1,1] and so does $\cos - \sin$, thus $f'$ is bounded. This makes $f$ a $Lipschitz$ function and so it is uniformly continuous.

All I'm asking is if this is correct and if not, could anyone give the proper solution? I'm asking this because the way I solved them seems to easy to be ok.

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    Everything is correct (anyway, *Lipschitz* has no 'n'). – 2017-02-01
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    @Crostul $f(1) = - \frac{1}{2}$?? – 2017-02-01
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    Since differentiable functions are continuous and the difference or sum of continuous functions is continuous, you can use the same argument for (b) as in (a)...... But if you were given (a) on a test or homework you might want to justify the continuity: $x^2-x-2=(x-2)(x+1)$ is continuous and not $0$ for $x\in [0,1]$ so $f$ is continuous.... Anyway it's all correct. – 2017-02-01

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