Answer:
Here's what I've done:
Let ϵ>0ϵ>0 be given.
Now,
|f(x)−f(y)|=|tan−1x−tan−1y|=∣∣tan−1(x−y1+xy)∣∣
|f(x)−f(y)|=|tan−1x−tan−1y|=|tan−1(x−y1+xy)|
For non-negative x,y∈R,x,y∈R,
|f(x)−f(y)|=|tan−1x−tan−1y|=∣∣tan−1(x−y1+xy)∣∣≤|tan−1(x−y)|≤|x−y|<δ
|f(x)−f(y)|=|tan−1x−tan−1y|=|tan−1(x−y1+xy)|≤|tan−1(x−y)|≤|x−y|<δ
We can choose ϵ=δϵ=δ. Now, for negative x,y∈Rx,y∈R. Please, how do I go about it?
