소단원 평가	Ⅰ. 함수의 극한과 연속			1. 함수의 극한							학년
	02. 함수의 극한에 대한 성질
계원예술고등학교		일시:			반:		번호:			이름:

1 두 함수 [eqn: f(x),~g(x)]에서

[eqn: lim fromx->1 f(x) =4], [eqn: lim fromx->1 g(x) = -3]일 때, 다음 극한값을 구하시오.

⑴ [eqn: lim fromx->1 left {f(x)+g(x) right }]	⑵ [eqn: lim fromx->1][eqn: left { 3f(x) - 2 g(x) right }]

⑶ [eqn: lim fromx->1][eqn: f(x) g(x)]

2 다음 극한값을 구하시오.

⑴ [eqn: lim fromx->1][eqn: (2x-3)]	⑵ [eqn: lim fromx->-1][eqn: (x^2 -5x+6)]

⑶ [eqn: lim fromx->2][eqn: (3x-1)(x+2)]	⑷ [eqn: lim fromx->-2][eqn: {x^2 -4x}over x+1]

3 다음 극한값을 구하시오.

⑴ [eqn: lim fromx->-1][eqn: {x^2 -4x-5}over x+1]      ⑵ [eqn: lim fromx->2][eqn: {rootx+2 -2}over x-2]

4 다음 극한값을 구하시오.

⑴ [eqn: lim fromx->3][eqn: x-3 over {x^2 -9 }]	⑵ [eqn: lim fromx->-1][eqn: {x^3 -x^2 -x+1}over{x^2 -1]

⑶ [eqn: lim fromx->4][eqn: {x^2 -4x}over{root x -2}]	⑷ [eqn: lim fromx->0][eqn: {root x+1 -1}over x]

5 다음 극한값을 구하시오.

⑴ [eqn: lim fromx->inf][eqn: {2x^2 -x+5}over{5x^2 -2x-4}]    ⑵ [eqn: lim fromx->inf][eqn: (root{x^2 +x} -x)]

6 다음 극한값을 구하시오.

⑴ [eqn: lim_x->inf][eqn: {2x^2 +3x-4}over{x^2 -x-3}]

⑵ [eqn: lim_x->inf][eqn: 1over{root{x^2 +3x}-x}]

7 다음 극한값을 구하시오.

⑴ [eqn: lim _{ x`->`-1 } {`` {x^2 -1} over {x+1} }]

⑵ [eqn: lim _{ x`->`-2 } { ``{x^2 +x-2} over {x+2}}]

8 함수 [eqn: f(x) = cases{ x^2 ~ & (x<1 ) # 3 over 2 & ( x=1 ) # x+1 & (x>1 ) }}]에서 [eqn: x->1]일 때, [eqn: f(x)]의 극한값을 구하시오.

9 다음 극한값을 구하시오

⑴ [eqn: lim fromx->-2][eqn: {x^2 -x- 6 } over x+2]	⑵ [eqn: lim fromx->0 x over {root 1+x - root 1-x}]

10 다음 극한값을 구하시오

⑴ [eqn: lim fromx->inf][eqn: {3x^2 -x+4}over{x^2 +1}]    ⑵ [eqn: lim fromx->inf][eqn: (root{x^2 + 9x} -x)]

02. 함수의 극한에 대한 성질

소단원 평가 정답 및 풀이

1 답 ⑴ [eqn: 1]   ⑵ [eqn: 18]    ⑶ [eqn: -12]

⑴ [eqn: lim fromx->1 left {f(x)+g(x) right }][eqn: = lim fromx->1 f(x)+ lim fromx->1 g(x)]	[eqn: = 4+(-3) = 1]

⑵ [eqn: lim fromx->1][eqn: left { 3f(x) - 2 g(x) right }][eqn: = 3 lim fromx->3 f(x) - 2 lim fromx->1 g(x)]

[eqn: = 3 times 4 -2 times (-3) =18]

⑶ [eqn: lim fromx->1][eqn: f(x) g(x)]	[eqn: = lim fromx->1 f(x) times lim fromx->1 g(x)]

[eqn: = 4 times (-3) =- 12]

2 답 ⑴ [eqn: -1]   ⑵ [eqn: 12]    ⑶ [eqn: 20]   ⑷ [eqn: -12]

⑴ [eqn: lim fromx->1][eqn: (2x-3)][eqn: = 2 lim fromx->1 x - lim fromx->1 3 = 2 times 1 -3 = ` -1]

⑵ [eqn: lim fromx->-1][eqn: (x^2 -5x+6)][eqn: =lim fromx->-1 x^2 - 5 lim fromx->-1 x + lim fromx->-1 6]

[eqn: = 1 - 5 times (-1) +6 = 12]

⑶ [eqn: lim fromx->2][eqn: (3x-1)(x+2)][eqn: = lim fromx->2 (3x^2 +5x -2)]

[eqn: = 3lim fromx->2 x^2 +5 lim fromx->2 x - lim fromx->2 2][eqn: = 3 times 4 +5 times 2 -2 = 20]

⑷ [eqn: lim fromx->-2][eqn: {x^2 -4x}over x+1][eqn: = {lim fromx->-2 (x^2 -4x) } over {lim fromx->-2 (x+1)}]

[eqn: = { lim fromx->-2 x^2 - 4 lim fromx->-2 x } over {lim fromx->-2 x+lim fromx->-2 1 }][eqn: = {4 - 4 times (-2) } over { -2+1} = -12]

3 답 ⑴ [eqn: -6]	⑵ [eqn: 1over4]

⑴ [eqn: lim fromx->-1][eqn: {x^2 -4x-5}over x+1][eqn: = lim fromx->-1][eqn: {(x+1)(x-5)}over x+1]

[eqn: = lim fromx->-1 (x-5)]

[eqn: = -1-5= -6]

⑵ [eqn: lim fromx->2][eqn: {rootx+2 -2}over x-2]

[eqn: = lim fromx->2 {(root x+2 -2 )( root x+2 +2 ) } over {(x-2)(root x+2 +2)}]

[eqn: = lim fromx->2 {x-2} over {(x-2)(root x+2 +2)}]

[eqn: = lim fromx->2 1 over {root x+2 +2}][eqn: = 1 over 2+2 = 1 over 4]

4 답 ⑴ [eqn: 1 over 6]   ⑵ [eqn: -2]    ⑶ [eqn: 16]   ⑷ [eqn: {1} over {2}]

⑴ [eqn: lim fromx->3][eqn: x-3 over {x^2 -9 }][eqn: = lim fromx->3 {x-3} over {(x-3)(x+3)}]

[eqn: = lim fromx->3 1 OVER {x+3} = 1 OVER 6]

⑵ [eqn: lim fromx->-1][eqn: {x^3 -x^2 -x+1}over{x^2 -1][eqn: =lim fromx->-1][eqn: {(x-1)^2 (x+1)}over{(x-1)(x+1)]

[eqn: =lim fromx->-1 (x-1)][eqn: = -1-1=-2]

⑶ [eqn: lim fromx->4][eqn: {x^2 -4x}over{root x -2}][eqn: = lim fromx->4 {x(x -4)(root x +2 )}over{(root x -2 ) (root x +2 )}]

[eqn: = lim fromx->4 {x(x -4)(root x +2 )}over{x-4 }]

[eqn: = lim fromx->4 {x(root x +2 )}]

[eqn: = 4 times (2+2) = 16]

⑷ [eqn: lim fromx->0][eqn: {root x+1 -1}over x]

[eqn: =lim fromx->0][eqn: {{(root x+1 -1)}(root x+1 + 1) } over {x(root x+1 +1) }]

[eqn: =lim fromx->0][eqn: {x } over {x(root x+1 +1) }]

[eqn: =lim fromx->0][eqn: {1} over {root x+1 +1}]

[eqn: = 1 over 1+1 = 1 over 2]

5 답 ⑴ [eqn: 2over5]	⑵ [eqn: 1over2]

⑴ [eqn: lim fromx->inf][eqn: {2x^2 -x+5}over{5x^2 -2x-4}] [eqn: = lim fromx->inf {2 - 1 over x + 5 over x^2 } over { 5 - 2 over x - 4 over x^2} = 2 over 5]

⑵ [eqn: lim fromx->inf][eqn: (root{x^2 +x} -x)]

[eqn: = lim fromx->inf { (root{x^2 +x} -x) (root{x^2 +x} +x)} over {root{x^2 +x} +x}][eqn: = lim fromx->inf { (x&^2 +x)-x^2 } over {root{x^2 +x} +x}][eqn: = lim fromx->inf { x } over {root{x^2 +x} +x}]

[eqn: = lim fromx->inf { 1 } over {root{1 + 1 over x } +1} = 1 over 2]

6 답 ⑴ [eqn: 2]	⑵ [eqn: {2} over {3}]

⑴ [eqn: lim_x->inf][eqn: {2x^2 +3x-4}over{x^2 -x-3}][eqn: = lim fromx->inf {2 + 3 over x - 4 over x^2 } over { 1 - 1 over x - 3 over x^2} = 2]

⑵ [eqn: lim_x->inf][eqn: 1over{root{x^2 +3x}-x}]

[eqn: = lim fromx->inf { root{x^2 +3x} +x} over {(root{x^2 +3x} -x) (root{x^2 +3x} +x) }][eqn: = lim fromx->inf { {root{x^2 +3x} + x} over { (x&^2 +3x)-x^2 }][eqn: = lim fromx->inf { {root{x^2 +3x} + x} over 3x][eqn: = lim fromx->inf {root{1 + 3 over x } +1} over 3 = 2 over 3]

7 답 ⑴ [eqn: -2]   ⑵ [eqn: {-3}]

⑴ [eqn: lim _{ x`->`-1 } { `` {x^2 -1} over {x+1}}][eqn: `=` lim _{ x`->`-1 } {`` { LEFT ( x+1 RIGHT ) LEFT ( x-1 RIGHT )} over {x+1} }]
[eqn: `=` lim _{ x`->`-1 } { `` LEFT ( x-1 RIGHT ) }`=` -2]

⑵ [eqn: lim _{ x`->`-2 } {`` {x^2 +x -2} over {x+2} }][eqn: `=` lim _{ x`->`-2 } {`` { LEFT ( x-1 RIGHT ) LEFT ( x+2 RIGHT )} over {x+2} }]
[eqn: `=` lim _{ x`->`-2 } {`` LEFT ( x-1 RIGHT ) }`=` -3]

8 답 없다.

[eqn: lim _{x -> 1-} {x ^{2}} =1], [eqn: lim _{x -> 1+} {(x+1)} =2]

[eqn: lim _{x -> 1-} {f(x)} != lim _{x -> 1+} {f(x)}]

따라서 [eqn: x->1]일 때 [eqn: f(x)]의 극한값은 없다.

9 답 ⑴ [eqn: -5]   ⑵ [eqn: 1]

⑴ [eqn: lim fromx->-2][eqn: {x^2 -x- 6 } over x+2][eqn: = lim fromx->-2 {(x+2)(x-3)} over {x+2}]

[eqn: = lim fromx->-2 (x-3) =` -5]

⑵ [eqn: lim fromx->0 x over {root 1+x - root 1-x}]

[eqn: = lim fromx->0 { x( root{1+x} +root 1-x )} over {( root{1+x} -root 1-x ) ( root{1+x} +root 1-x ) }][eqn: = lim fromx->0 { x( root{1+x} +root 1-x )} over {2x }][eqn: = lim fromx->0 { root{1+x} +root 1-x } over {2 }][eqn: =1]

10 답 ⑴ [eqn: 3]   ⑵ [eqn: 9 over 2]

⑴ [eqn: lim fromx->inf][eqn: {3x^2 -x+4}over{x^2 +1}][eqn: = lim fromx->inf {3 - 1 over x + 4 over x^2 } over { 1 + 1 over x^2} = 3]

⑵ [eqn: lim fromx->inf][eqn: (root{x^2 + 9x} -x)]

[eqn: = lim fromx->inf { (root{x^2 +9x} -x) (root{x^2 +9x} +x)} over {root{x^2 +9x} +x}][eqn: = lim fromx->inf { (x&^2 +9x)-x^2 } over {root{x^2 +9x} +x}][eqn: = lim fromx->inf { 9x } over {root{x^2 +9x} +x}]

[eqn: = lim fromx->inf { 9 } over {root{1 + 9 over x } +1} = 9 over 2]