Lim x->1 of x^2=1
Lim x->1 of sin(1/(x-1)) -> sin(∞)
The latter is undefined because of the oscillatory nature of sin(x).
So the required limit is undefined.
Similar Questions
1. Explain why the functionf(x)=(x^2-4)/(x-2) is not continous on [0,3]. what kind of discontinuity occurs? 2. use areas to show
1 answer
Evaluate the limit:Limit as x approaches 6 from the right: Sq.root of (x - 6). I know the limit is 0, but how do I show this?