# 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? Relevance
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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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• (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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• A is correct. See graph below. • Thank you.

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