Math 220 March 7 I. If 1200 cm2 of material is available to make a ...

12.
13.
Answer:
Answer:
Answer:
lim (sin(x) ln(x))
x→0 +
= lim
x→0 +
ln(x)
csc(x)
L.H.
= lim
x→0 +
1/x
− csc(x) cot(x)
= lim
x→0 +
− sin2
(x) 0
x cos(x) 0
L.H.
= lim
x→0 +
−2 sin(x) cos(x)
cos(x) − x sin(x)
= 0
1
= 0
lim
x→0 +
2
x − ln(x )
lim
x→0 +
2
x − ln(x )
= lim (x − 2 ln(x))
x→0 +
= 0 − (−∞)
= ∞
lim (tan(3x))6x
x→0 +
12