一道求取值范围的题目

f(r) = 10r/(40+5r)
= 2 - 80/(40+5r)
f'(r) = 80/(40+5r)^2 >0 ( r≠ -8)

lim(r->∞) 10r/(40+5r) = 2
取值范围=(2, ∞ )追问

抱歉，我看不懂那个lim……

而且不太懂第一步

追答

10r = 2(40+5r) -80
f(r) = 10r/(40+5r)
= 2 - 80/(40+5r)
f'(r) = 400/(40+5r)^2 >0 ( r≠ -8)

lim(r->∞) 10r/(40+5r)
=lim(r->∞) 10/(40/r+ 5)
=10/5
= 2

lim(r->-8+) 10r/(40+5r) -> -∞
=lim(r->-8-) 10/(40/r+ 5)->∞
取值范围=(-∞ , 2 ) U (2, ∞ )

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