How do you evaluate # (7r + 3)/(5r^2 - 2r + 2) # as r approaches infinity?

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Answer 1 · 2016-06-21T05:26:20

#0#

As we see, the highest power in the whole fraction is 2 (in #5r^2#), so we divide both numerator and denominator by #r^2#

#((7/r)+(3/r^2))/(5-(2/r)+(2/r^2))#

As r approaches infinity, #1/r# approaches zero

Hence, substituting 0 for #1/r#, we get

#(0+0)/(5-0+0)#

This becomes #0/5# which is #0#

Hence the answer is #0#