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Evaluate the limit using L'Hospital's rule if necessary limit ((9x)/(9x+3))^(7x) x->inf
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This one looks like a toughie. First we can factor 3 out of the numerator and denominator: ((9x)/(9x+3))^(7x) = (3x/(3x+1))^(7x) Let y = ((3x)/(3x+1))^(7x) Take natural logarithm on both sides: ln(y) = (7x) * ln(3x/(3x+1)) See if you can take the limit as x->infinity. 7 is a constant that can be factored out of the limit.
|dw:1383879292946:dw| idk if this is right
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