Calculus 1 problem: functions?
Im doing some older exams that my professor has provided, but I havent got the solutions for this one. Can someone help confirm that the solution Ive arrived at is correct?
- Ray SLv 71 month ago
A function ƒ is continuous at x=a if:
𝟏. lim ƒ(x) exists
𝟐. ƒ(𝓪) exists
𝟑. lim ƒ(x) = ƒ(𝓪)
Since (ℯˣ -1) / x is 0/0 at x = 0, lim (ℯˣ -1) / x as x➔0 can be
determined by using L'Hopital's Rule.
Therefore, lim( (ℯˣ-1)/x ) ＝ lim( ℯˣ/1 ) ＝ ℯ⁰ ＝ 1
so that defining the point (0,1) fills the hole in the graph of
y = (ℯˣ -1) / x making g(x) continuous at x = 0.
- fcas80Lv 71 month ago
(e^x - 1)/x is undefined at x = 0. It is a 0/0 expression. But by L'Hospital's Rule, its limit is 1. And further, the function g is defined to be 1, although this is not needed. So yes, it is continuous.
- billrussell42Lv 71 month ago
A is correct. See graph below.