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30<br />
8x−5 8x−5<br />
Εποµένως, lim<br />
= lim<br />
=<br />
x→−∞<br />
2<br />
x→−∞<br />
x + 2x+ 4 − x+<br />
3<br />
2 4<br />
| x| 1+ + − x+<br />
3<br />
2<br />
x x<br />
⎛ 5 ⎞<br />
x 8 −<br />
5<br />
8 −<br />
8x − 5<br />
⎜ ⎟<br />
x<br />
lim = lim<br />
⎝ ⎠<br />
= lim x =<br />
x→−∞ 2 4<br />
x→−∞ ⎛<br />
x<br />
2 4 3⎞<br />
→−∞<br />
2 4 3<br />
− x 1+ + − x+ 3 1 1<br />
2<br />
x⎜− 1+ + − 1+<br />
2 ⎟ − + + − +<br />
2<br />
x x<br />
⎝ x x x⎠<br />
x x x<br />
5<br />
8−<br />
lim<br />
x→−∞<br />
x<br />
8−<br />
0<br />
= = =−4<br />
.<br />
2 4 3 − 1+ 0+ 0 − 1+<br />
0<br />
− 1+ lim + lim − 1+<br />
lim<br />
x→−∞ x<br />
2<br />
x →−∞ x x→−∞<br />
x<br />
⎛ 1 ⎞ 1<br />
x −1 1<br />
−1 lim −1<br />
x −x ⎜ ⎟<br />
x x<br />
0 1<br />
iii) lim lim lim<br />
x →+∞ x −<br />
=<br />
⎝ ⎠<br />
= = = =−1<br />
.<br />
x→+∞ x −2 x→+∞ ⎛ 2 ⎞ x→+∞<br />
2 2<br />
1 1 lim<br />
1−0<br />
x⎜1−<br />
⎟ − −<br />
x<br />
x<br />
x→+∞<br />
⎝ ⎠<br />
x<br />
3 3<br />
2 | x | 1+ 1+<br />
2 2<br />
x + 3 1 0<br />
iv) lim lim<br />
x<br />
lim<br />
x +<br />
= =− =− =−1<br />
.<br />
x→−∞ x −4 x→−∞ ⎛ 4 ⎞ x< 0 x→−∞<br />
4<br />
1<br />
1−0<br />
x⎜1−<br />
⎟<br />
−<br />
⎝ x ⎠<br />
x<br />
2 2<br />
2 | x | 1+ −x − x 1+ −x<br />
2 2<br />
x + 2 −x v) lim = lim<br />
x<br />
= lim<br />
x<br />
=<br />
x→−∞ 2 x→−∞ 1<br />
x< 0 x→−∞<br />
9x<br />
+ 1+<br />
x<br />
1<br />
| x | 9+ + x − x 9+ + x<br />
2 2<br />
x<br />
x<br />
2<br />
− 1+ −1<br />
2<br />
1 0 1<br />
lim<br />
x − + −<br />
= = = 1 .<br />
x→−∞<br />
1 − 9+ 0 + 1<br />
− 9+ + 1<br />
2<br />
x<br />
2<br />
⎛ x − 3x+ 1 ⎞<br />
3. Να βρεθούν τα ab∈R , , ώστε lim ⎜ − ax − b⎟<br />
= 0 .<br />
x→+∞<br />
⎝ 2x<br />
−1<br />
⎠<br />
2<br />
⎛ x 3 1<br />
Λύση: Εφόσον lim − x + 2<br />
⎞<br />
⎜ − ax − b⎟<br />
= 0 και<br />
x→+∞<br />
⎝ 2x<br />
−1<br />
lim ⎛ x − 3x+ 1 b⎞<br />
⎜<br />
a<br />
x→+∞<br />
2 − − ⎟ =<br />
⎠<br />
⎝ 2 x − x x⎠<br />
2<br />
⎛1⎛<br />
x − 3x+<br />
1 ⎞⎞<br />
= lim ax b 0<br />
x→+∞⎜<br />
⎜ − − ⎟<br />
=<br />
x 2x−1<br />
⎟<br />
.<br />
⎝ ⎝<br />
⎠⎠<br />
2 2<br />
⎛ x − 3x+ 1 b⎞<br />
x − 3x+<br />
1 b 1<br />
Άρα 0 = lim ⎜ −a− lim a lim a<br />
x<br />
2 ⎟= − − = −<br />
→+∞<br />
2<br />
2x x x x→+∞ 2x x x→+∞<br />
⎝ − ⎠<br />
−<br />
x 2<br />
1<br />
a = .<br />
2<br />
2 2<br />
⎛ x − 3x+ 1 1 ⎞ ⎛ x − 3x+<br />
1 1 ⎞<br />
Τώρα, lim ⎜ − x − b⎟= 0 ⇔ b= lim ⎜ − x⎟=<br />
x→+∞<br />
2x<br />
1 2 x→+∞<br />
⎝ −<br />
⎠ ⎝ 2x−1 2 ⎠<br />
2<br />
2x − 6x+ 2 −x(2x−1) − 5x+<br />
2 5<br />
= lim<br />
= lim =− .<br />
x→+∞<br />
2(2x−1) x→+∞<br />
4x−2 4<br />
Η γεωµετρική ερµηνεία του παραπάνω γεγονότος είναι η ακόλουθη:<br />
και εποµένως,