lim ΔTC/Δq and TC= 2q^2 + 5q + 5
how calculate lim 2q^2 + 5q + 5/ Δq
Δq→0 ?
thanx
2007-10-11 03:30:10 · 3 answers · asked by aurelijad 1 in Science & Mathematics ➔ Mathematics
this limit is dTC/dq !
so dTC/dq= 4q+5
2007-10-11 03:52:41 · answer #1 · answered by Anonymous · 2⤊ 0⤋
TC(q) = 2q^2 + 5q + 5
So TC(q+Δq) = 2(q+Δq)^2 + 5(q+Δq) + 5
You are asked to compute [TC(q+Δq) - TC(q)]/Δq
and then take the limit as Δq→0
Before taking that limit, you will need to expand out TC(q+Δq) and subtract TC(q). If you've done it correctly, you will be able to factor out a Δq, with which you can cancel the Δq in the denominator, and then take the limit as Δq→0.
2007-10-11 04:00:17 · answer #2 · answered by Anonymous · 0⤊ 0⤋
lim ΔTC/Δq (2q^2 +5q+5)
Δq→0
= lim [2(q+Δq)^2 +5(q+Δq) +5 -(2q^2+5q+5)]/Δq
Δq→0
= lim [2q^2 +4qΔq +2Δq^2 +5q +5Δq +5 -2q^2-5q-5]/Δq
Δq→0
= lim [4qΔq +2Δq^2 +5Δq ]/Δq
Δq→0
= lim [4q +2Δq +5]
Δq→0
= 4q +5
2007-10-11 03:57:47 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋