How to proceed for the following problem in uniform continuity

Question

I am trying to solve this problem. Please give one hint to proceed for the problem. I think there may be counterexample

Answer 1

Yes, this is true. I'm not sure how to give a hint that does not spoil it. I suggest reading sentence by sentence and going for your self once you think you can.

Since $f$ is uniformly continuous, by definition: $\forall \varepsilon > 0 \exists \delta > 0$ s.t. $$\forall x,y \in (0, \infty): \vert x - y \vert < \delta \Rightarrow \vert f(x) - f(y) \vert < \varepsilon$$. Hence in particular this is true for $y := x + \frac{1}{x}$. Hence for any given $\varepsilon > 0$ there exists $\delta > 0$ with

$$\vert \frac{1}{x}\vert = \vert x - x - \frac{1}{x} \vert < \delta \Rightarrow \vert f(x) - f(x + \frac{1}{x}) \vert < \varepsilon$$ Hence for $x \rightarrow \infty$ we have $\vert f(x) - f(x + \frac{1}{x}) \vert \rightarrow 0$ and thus

$$\lim_{x \rightarrow \infty} \frac{f(x+\frac{1}{x})}{f(x)} = \lim_{x \rightarrow \infty} \frac{f(x) +( f(x+\frac{1}{x})-f(x))}{f(x)} = 1 + \lim_{x \rightarrow \infty} f(x+\frac{1}{x})-f(x)) = 1$$