Page 10 - e-book Calculcs I
P. 10
3
12. lim →−∞ 3 √ + 2 − 1
1
lim →−∞ 3 √ + 2 − 1 = lim x→−∞ 3 √x (1 + x 2 2 − )
3
3
3
x
3
= lim →−∞ √1 + 2 2 − 1 3
= -∞
2
13. lim →0 1−
2
2
lim →0 1− = lim →0 1− ∙ 1+ 5
1+ 5
2
= lim →0 (1+ 5 )
2
1−
1
= lim →0 ∙ ∙ (1 + 5 ) ∙ ∙
= 2
14. lim →∞ arcsec =
arcsec =
sec = ∞
1
= ∞
cos
= 0
=
2
lim →−∞ =
arcsec =
sec = −∞
1
= −∞
cos
= 0
=
2
15. lim →0 + 1+
0
lim →0 + 1+ = lim →0 + 1+ = = 0
2
+
( ) > 0 ; → 0
1 +
∴ lim = +∞
→0 +
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