# 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?

### 3 Answers

- Ray SLv 71 month ago
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A function ƒ is continuous at x=a if:

𝟏. lim ƒ(x) exists

𝐱➔𝓪

𝟐. ƒ(𝓪) exists

𝟑. lim ƒ(x) = ƒ(𝓪)

𝐱➔𝓪

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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

x➔0 x➔0

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.

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- 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.

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- billrussell42Lv 71 month ago
A is correct. See graph below.

- Robin1 month agoReport
Thank you.

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