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Here we prove that if f(r) = e then f'(r) = f(r). eh - 1 eh – 1 (a) Consider lim Explain why, if we let y = e -1 then lim lim yo In(y+ 1)* Be sure to justify h-0 h h h0 the change from h 0 to y +0. 1 (b) Show, step by step, that lim y40 In(y + 1) lim In(1 + y)'/v 1 1 (c) Show, step by step, that lim In(1+ y)'/v In (lim (1+ 4)") d = e dr (d) Use the above to prove that

Here we prove that if f(r) = e then f'(r) = f(r). eh - 1 eh – 1 (a) Consider lim Explain why, if we let y = e -1 then lim lim yo In(y+ 1)* Be sure to justify h-0 h h h0 the change from h 0 to y +0. 1 (b) Show, step by step, that lim y40 In(y + 1) lim In(1 + y)'/v 1 1 (c) Show, step by step, that lim In(1+ y)'/v In (lim (1+ 4)") d = e dr (d) Use the above to prove that

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Here we prove that if \( f(x) = e^x \) then \( f'(x) = f(x) \). (a) Consider \( \lim_{h \to 0} \frac{e^h - 1}{h} \). Explain why, if we let \( y = e^h - 1 \) then \( \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{y \to 0} \frac{y}{\ln(y+1)} \). Be sure to justify the change from \( h \to 0 \) to \( y \to 0 \). (b) Show, step by step, that \( \lim_{y \to 0} \frac{y}{\ln(y+1)} = \frac{1}{\lim_{y \to 0} \ln(1+y)^{1/y}} \). (c) Show, step by step, that \( \frac{1}{\lim_{y \to 0} \ln(1+y)^{1/y}} = \ln \left( \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \right) = 1 \). (d) Use the above to prove that \( \frac{d}{dx} e^x = e^x \).
Transcribed Image Text:Here we prove that if \( f(x) = e^x \) then \( f'(x) = f(x) \). (a) Consider \( \lim_{h \to 0} \frac{e^h - 1}{h} \). Explain why, if we let \( y = e^h - 1 \) then \( \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{y \to 0} \frac{y}{\ln(y+1)} \). Be sure to justify the change from \( h \to 0 \) to \( y \to 0 \). (b) Show, step by step, that \( \lim_{y \to 0} \frac{y}{\ln(y+1)} = \frac{1}{\lim_{y \to 0} \ln(1+y)^{1/y}} \). (c) Show, step by step, that \( \frac{1}{\lim_{y \to 0} \ln(1+y)^{1/y}} = \ln \left( \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \right) = 1 \). (d) Use the above to prove that \( \frac{d}{dx} e^x = e^x \).
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