Proving Continuity of exp(x) at c=0

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SUMMARY

The discussion focuses on proving the continuity of the exponential function, exp(x), at c=0 using delta-epsilon definitions. The key approach involves demonstrating that |exp(x) - exp(0)| = |exp(x) - 1| can be made less than ε by appropriately choosing δ. The participant suggests setting δ = ln(1 + ε) and references two important inequalities: (a) lim_n->inf y^(1/n) = 1 for y > 0 and (b) exp(x) < exp(y) for x < y. The proof requires leveraging these inequalities without using logarithmic functions.

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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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