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Unformatted text preview: The new expression inside the limit as R approaches π /2, is lnsec(R) + tan(R) + lncos(R)  ( ln1 + ln1). The value of ( ln1 + ln1) = 0, so this part is not necessary for extra computation. Using the logarithm properties, the lnsec(R) + tan(R) + lncos(R) can be rewritten as ln sec(R) + tan(R) * cos(R), continuing to take the limit as R approaches π /2. Because the sec(x) = 1/cos(x), and because the definition of tan(x) = sin(x)/cos(x), the logarithmic expression can be simplified to ln1+sin(R) as the limit of R approaches infinity, by multiplying sec(R) and tan(R) to the cos(R). Finally, the limit as R approaches π /2, should be evaluated. This can easily be done if π /2 is substituted for R. This produces the expression ln1 + sin( π /2). The sin( π /2) = 1, so the solution to the problem is ln1 + 1 = ln2....
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This note was uploaded on 02/29/2012 for the course MATH 151 taught by Professor Sc during the Fall '08 term at Rutgers.
 Fall '08
 sc
 Calculus

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