exp_continuity_1013

# exp_continuity_1013 - THE CONTINUITY OF y = ex Here we show...

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y = e x Here we show that y = f ( x ) = e x is continuous at x 0 for any real number x 0 . To do so, we must show that lim x x 0 f ( x ) = lim x x 0 e x is equal to f ( x 0 ) = e x 0 . Let’s deFne Δ x by x = x 0 + Δ x (so x x 0 is equivalent to Δ x 0), and use the fact that e x 0 x = e x 0 e Δ x . Then what we are left to show is that lim x x 0 h e x 0 e Δ x i = e x 0 . But x 0 is held constant as x x 0 , so the factor e x 0 can be taken outside the limit (by the constant-multiple limit theorem), leaving us with the problem of showing that e x 0 ± lim x x 0 h e Δ x = e x 0 or, equivalently, that lim x x 0 h e Δ x i = 1 , (1) where the lim x x 0 is equivalent to lim Δ x 0 . We will prove this last relation using the algebraic ( ² - δ ) formulation of the limit problem discussed in class, and in the textbook. PROOF OF EQUATION (1) We Frst set the condition of closeness of the function, e Δ x , to the claimed limit, 1, using ² > 0 as the measure of this degree of closeness:

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exp_continuity_1013 - THE CONTINUITY OF y = ex Here we show...

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