Tolong di bantun pleaSe​

By Posted on

Tolong di bantun pleaSe​

Tolong di bantun pleaSe​

⟩ Limit Fungsi Aljabar

.

a) 4

b) –¼

.

» Penyelesaian

.

[A]

displaystyle{rm lim limits_{x to : 3} frac{ {x}^{2} + 2x - 15 }{ {x}^{2} - 4x + 3}}

displaystyle{rm lim limits_{x to : 3} frac{ (x + 5) cancel{(x - 3)} }{ cancel{(x - 3)}(x - 1)} }

displaystyle{rm lim limits_{x to : 3} frac{ (x + 5) }{ (x - 1)}}

displaystyle{rm = frac{ (3 + 5) }{ (3 - 1)}}

displaystyle{ rm = frac{ 8 }{ 2}}

displaystyle{rm = 4}

.

[B]

displaystyle{rm lim limits_{x to : 3} frac{ 2 - sqrt{x + 1} }{ x - 3}}

displaystyle{rm lim limits_{x to : 3} frac{ 2 - sqrt{x + 1} }{ x - 3} . frac{2 + sqrt{x + 1} }{2 + sqrt{x + 1} }}

displaystyle{rm lim limits_{x to : 3} frac{ 4 - ( {x + 1)} }{ (x - 3)(2 + sqrt{x + 1)}} }

displaystyle{rm lim limits_{x to : 3} frac{ 4 - {x - 1} }{ (x - 3)(2 + sqrt{x + 1)} }}

displaystyle{rm lim limits_{x to : 3} frac{ - {x + 3} }{ (x - 3)(2 + sqrt{x + 1)} }}

displaystyle{rm lim limits_{x to : 3} frac{ - 1 cancel{(x - 3)}}{ cancel{ (x - 3)}(2 + sqrt{x + 1)} }}

displaystyle{rm lim limits_{x to : 3} frac{ - 1 }{ (2 + sqrt{x + 1)} }}

displaystyle{ = rm frac{- 1 }{ (2+ sqrt{3 + 1)} }}

displaystyle{= rm frac{- 1 }{ 2+ sqrt{4} }}

displaystyle{ = rm frac{- 1 }{ 2 + 2 }}

displaystyle{= rm - frac{1 }{4}}