F(x)=(e^1/x +1)/(e^1/x -1) is function continuous at x=0?
Do you mean #F(x)=(e^(1/x) +1)/(e^(1/x) -1)#
Do you mean
2 Answers
No. Lateral limits are different at
Explanation:
so
- because
and with rules De L'Hopital
(Note: you can check in the graph the behaviour of
graph{e^x [-22.79, 22.82, -11.42, 11.37]}
Explanation:
Both
both
Because
the quotient