Prove f(x)=1xf(x) = \frac{1}{x}f(x)=x1 is NOT uniformly continuous on (0,1)(0,1)(0,1). Which argument works?
Take xn=1/nx_n = 1/nxn=1/n and yn=1/(n+1)y_n = 1/(n+1)yn=1/(n+1); then ∣xn−yn∣→0|x_n - y_n| \to 0∣xn−yn∣→0 but ∣f(xn)−f(yn)∣=1↛0|f(x_n)-f(y_n)| = 1 \not\to 0∣f(xn)−f(yn)∣=1→0
fff is discontinuous at x=0x=0x=0
f′(x)=−1/x2f'(x) = -1/x^2f′(x)=−1/x2 is unbounded, so fff cannot be uniformly continuous
Both (a) and (c) are valid arguments