How do you find the limit of #(x^2+5x-7)/(e^(2x))# as x approaches #oo#?

1 Answer
Jul 2, 2016

#=0#

Explanation:

#lim_{x to oo} (x^2+5x-7)/(e^(2x))#

by inspection, the dominant term is in the denominator, ie the exponent, so the answer seems obvious, but we can do some more stuff to explore the obvious.....!

the immediate limit is #oo/oo# ie indeterminate so we can use L'Hopitals Rule

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# \lim _{x\to c} \frac {f(x)}{g(x)}= #

#\lim _{x\to c}\frac {f'(x)}{g'(x) }#

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#lim_{x to oo} (x^2+5x-7)/(e^(2x))#

#= lim_{x to oo} (2x+5)/(2e^(2x))#

#= lim_{x to oo} (2)/(4e^(2x))#

#= (2)/ {lim_{x to oo} (4e^(2x))}#

#= 2/oo#

#=0#