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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4.<br />
5.<br />
Answer:<br />
Answer:<br />
Answer:<br />
lim<br />
x→0 +<br />
ln( √ x)<br />
x2 = lim<br />
x→0 +<br />
(1/2) ln(x)<br />
= lim<br />
x→0 +<br />
= lim<br />
x→0 +<br />
= −∞<br />
5<br />
lim<br />
x→0<br />
x − 7x x 2<br />
ln(x)<br />
2x2 1 1<br />
(ln(x) − ) (∞ · −∞)<br />
x 2x<br />
5<br />
lim<br />
x→0<br />
x − 7x x<br />
x<br />
L.H.<br />
= lim<br />
x→0<br />
<br />
0<br />
0<br />
= ln(5) − ln(7)<br />
ln(5)5 x − ln(7)7 x<br />
1<br />
sin(x) − x<br />
lim<br />
x→0 x<br />
<br />
sin(x) − x 0<br />
lim<br />
x→0 x 0<br />
L.H. cos(x) − 1<br />
= lim<br />
x→0 1<br />
= 0<br />
9
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