This document contains 7 limit problems worked out step-by-step with explanations of the reasoning. The problems involve various algebraic expressions containing polynomials, rational expressions, and variables approaching different limits. The solutions find the limits by simplifying the expressions, factorizing where possible, and evaluating the behavior of terms as the variable approaches the limit.

1. 3x − 2
1. lim
x →+∞ 9x + 7
= ;
Απάντηση :
2
3−
3x − 2 x = 3−0 = 1
lim = lim
x →+∞ 9 x + 7 x →+∞ 7 9+0 3
9+
x
6 x 2 + 2 x +1
2. lim
x →−∞ 6 x 2 − 3 x + 4
=;
Απάντηση :
2 1 2 1
x 2 (6 + + 2 ) 6+ + 2
6 x 2 + 2 x +1 x x = lim x x = 6 =1
lim = lim
x →−∞ 6 x 2 − 3 x + 4 3 4 3 4
6− + 2 6
x →−∞ 2 x →−∞
x (6 − + 2 )
x x x x
3. lim ( x 2 +1 − x )
x→+∞ =;
Απάντηση :

2. lim( x 2 − 4 x + 1) = lim x 2 − lim 4 x + 1 = 4 − 8 + 1 = −3
x→2 x→2 x→ 2
x 2 − 4 lim( x − 4)
2
( x − 4)( x + 4) x−4
lim = x →2
= lim = lim =0
x→2 x + 4 lim( x + 4) x→2 x+4 x →2 1
x →2
lim (3x 2 + x ) = lim (3x 2 ) + lim x = 3 lim x 2 + lim x =
x →+∞ x →+∞ x →+∞ x →+∞ x →+∞
3(+∞) + ( +∞) = ( +∞) + ( +∞) = +∞
2
4. lim ( x 2 + 2 + x) =;
x →−∞
Απάντηση :
( x 2 + 2 + x)( x 2 + 2 − x) x 2 + 2 − x2 2
lim ( x 2 + 2 + x ) = lim = lim = lim =
x →−∞ x →−∞
( x 2 + 2 − x) x →−∞ 2 x →−∞ 2
x (1 + 2 ) − x
2
−x 1 + 2 − x
x x
2  1  2 2
lim = lim  ÷ lim = 0. =0
x →−∞ − x x →−∞ 1 + 0 +1
x →−∞ 2   2
− x[ 1 + + 1] 1 + 2 +1
x2 x
5. lim( x 2 − 4 x +1) =;
x →2
Απάντηση :
lim( x 2 −4 x + =lim x 2 −
1) lim 4 x + =4 − + =−
1 8 1 3
x→2 x→2 x→2

3. x2 − 4
6. lim = ;
x→2 x+4
Απάντηση :
x 2 − 4 lim( x − 4) lim x − 4 4 − 4
2 2
lim = x →2 = x →2 = =0 ή
x →2 x +4 lim( x + 4) lim x + 4 2 +4
x →2 x →2
x 2 − 4 lim( x − 4)
2
( x − 4)( x + 4) x −4
lim = x →2
= lim = lim =0
x →2 x + 4 lim( x + 4) x →2 x +4 x →2 1
x →2
7. lim (3 x 2 + x) =
x →+∞
Απάντηση :
lim (3 x 2 + x ) = lim (3 x 2 ) + lim x = 3 lim x 2 + lim x =
x →+∞ x →+∞ x →+∞ x →+∞ x →+∞
3( +∞) 2 + ( +∞) = ( +∞) + ( +∞) = +∞