SolutionLimitProbWithExplanation

- 7x 2 x→∞ 5x The following uses the natural log function L’Hopital’s Rule and no ”tricks” Find the following limit lim Let y = lim

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Unformatted text preview: 7x 2 x→∞ 5x The following uses the natural log function, L’Hopital’s Rule, and no ”tricks”. Find the following limit: lim Let y = lim x→∞ 2 1+ 5x 1+ 7x Now use the properties of the natural log function to write the exponential expression as a product. Remember, the natural log function is continuous on (0, ∞). ln y = ln lim x→∞ 2 1+ 5x 7x 7x 2 = lim ln 1 + x→∞ 5x We can write our product as a quotient by dividing by 2 ln y = lim 7x ln 1 + x→∞ 5x = lim 7x ln 1 + x→∞ 2 5x 1 instead of multiplying by 7x . 7x ln 1 + 2 5x = lim x→∞ 1 7x Now we have a quotient, so we can see if we can use L’Hopital’s Rule. 0 This is a ” ” form (remember, ln(1) = 0) so we can use L’Hopital’s Rule. 0 Note 1. Remember to use the chain rule for the derivative of ln 1 + Note 2. For ease of differentiation, we can write d2 = dx 5x 2 5 ln 1 + So 2 5x lim x→∞ 1 7x −1 x2 and = lim x→∞ So now we have ln y = ln lim d1 = dx 7x 1 1 + 52 x 1 7 x→∞ 2 1+ 5x 2 = 5x 1 7 = 1 x and 1 = 7x 1 7 1 x −1 . Using L’Hopital’s Rule, we get: x2 −1 2 5 x2 −1 x2 7x 2 5 2 . 5x 1 1 + 52 x 1 7 = lim x→∞ 14 . 5 1+ Finally, taking the exponential we get y = lim x→∞ 1 2 5x 7x 14 =e5. 2 5 2 5 1 = = 1 7 14 5 ...
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This note was uploaded on 09/25/2011 for the course CALCULUS 135 taught by Professor Augustarainsford during the Spring '11 term at Rutgers.

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