Proving Continuity of exp(x) at c=0

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To prove that e^x is continuous at c=0 using the delta-epsilon definition, the focus is on showing that for every ε > 0, there exists a δ > 0 such that |exp(x) - exp(0)| < ε whenever |x - 0| < δ. The expression simplifies to |exp(x) - 1| < ε, leading to the condition exp(x) < 1 + ε. By utilizing the limit property from the homework equations, specifically that as n approaches infinity, y^(1/n) approaches 1, one can establish a suitable δ. The suggestion is to set δ based on the natural logarithm, but the emphasis is on deriving δ using the conditions provided without directly using logarithms. This approach effectively demonstrates the continuity of e^x at c=0.
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Homework Statement


use delta, epsilon to prove that e^x is continuous at c = 0


Homework Equations



(a) for y>0, lim_n-> inf, y^(1/n) = 1
(b) for x < y, exp(x) < exp(y)

The Attempt at a Solution



im not sure how to approach this problem.
i have,
|exp(x) - exp(0)|= |exp(x) - 1|
so then exp(x) < 1 + ε
for δ > 0,
exp(δ) < 1 + ε

so then, i would set δ=ln(1+ε) for the proof?

also I am not sure how to use relevant eqn (a) to help with the problem. some insight would be appreciated. thank you.
 
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You are not supposed to use log; you are supposed to use the two conditions (a) and (b); first you realize that the limit in (a) is fairly close to your limit, if you set y=e, then you can use that limit to come up with a delta which is a function of the N(epsilon) of that limit.
 
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