Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
Math 220 March 7 I. If 1200 cm2 of material is available to make a ...
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14.<br />
Answer:<br />
ln(y) = ln( lim<br />
x→0 + (tan(3x))6x )<br />
= lim 6x ln(tan(3x)) (0(−∞))<br />
x→0 +<br />
= lim<br />
x→0 +<br />
6 ln(tan(3x))<br />
x −1<br />
−∞<br />
∞<br />
<br />
L.H.<br />
= lim<br />
x→0 +<br />
18 sec2 (3x)<br />
tan(3x)<br />
−x−2 = lim<br />
x→0 +<br />
−18x2 sec2 (3x)<br />
tan(3x)<br />
= lim<br />
x→0 +<br />
−18x2 cos(3x)<br />
sin(3x) cos2 (3x)<br />
= lim<br />
x→0 +<br />
−18x2 <br />
0<br />
sin(3x) cos(3x) 0<br />
L.H.<br />
= lim<br />
x→0 +<br />
−36x<br />
cos2 (3x) − sin2 (3x)<br />
= 0<br />
1<br />
= 0<br />
y = e ln(y)<br />
= e 0<br />
= 1<br />
lim (x)2/(1−x)<br />
x→1 +<br />
13
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