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Originally Posted by redsox25 Any positive number raised to any power is always positive ==> the limit cannot be a negative number. When x --> oo, e^(-3x) --> 0 and e^5x --> oo, so you can't uise L'Hospital as you did. The limit is e^(-3x)*1/(e^(5x)) --> 0*0 = 0 Tonio
Originally Posted by redsox25 you can't use L'Hopital in this case. numerator goes to , denominator goes to try re-writing the expression ...
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