Mohon jawab dengan cara nya ya

Mohon jawab dengan cara nya ya

Jawab:

limit bentuk 0/0

$sf h). ~lim_{xto -5}~~ dfrac{5x^2+ 26x + 5}{x^4 -8x^3 -63x^2}$

$sf ~lim_{xto -5}~~ dfrac{(x+ 5)(5x + 1)}{x^2(x + 5)(x -13)}$

$sf ~lim_{xto -5}~~ dfrac{(5x + 1)}{x^2(x -13)}$
sub x = – 5

$sf ~lim_{xto -5}~~ dfrac{(5(-5) + 1)}{(-5)^2(-5 -13)} = dfrac{-25 +1}{25(-18)}$

$sf ~lim_{xto -5}~~ dfrac{-24}{25(-18)} = dfrac{4}{75}$

$sf i. ~lim_{xto 3}~~ dfrac{9- x^2}{2sqrt{x^2 + 3} - 4sqrt3}$

$sf ~lim_{xto 3}~~ dfrac{(9- x^2)(2sqrt{x^2 + 3} + 4sqrt3)}{(2sqrt{x^2 + 3} - 4sqrt3)(2sqrt{x^2 + 3} + 4sqrt3) }$

$sf ~lim_{xto 3}~~ dfrac{(9- x^2)(2sqrt{x^2 + 3} + 4sqrt3)}{4(x^2+3) - 48}$

$sf ~lim_{xto 3}~~ dfrac{(9- x^2)(2sqrt{x^2 + 3} + 4sqrt3)}{4x^2+12 - 48}$

$sf ~lim_{xto 3}~~ dfrac{(9- x^2)(2sqrt{x^2 + 3} + 4sqrt3)}{4x^2 - 36}$

$sf ~lim_{xto 3}~~ dfrac{(9- x^2)(2sqrt{x^2 + 3} + 4sqrt3)}{-4(9 -x^2)}$

$sf ~lim_{xto 3}~~ dfrac{(2sqrt{x^2 + 3} + 4sqrt3)}{-4}$

x = 3

$sf ~lim_{xto 3}~~ dfrac{(2sqrt{9 + 3} + 4sqrt3)}{-4}$
$sf ~lim_{xto 3}~~ dfrac{(4sqrt3+ 4sqrt3)}{-4} = dfrac{8sqrt3}{-4} = - 2sqrt 3$

$sf j. ~lim_{xto 4}~~ dfrac{6x sqrt x - 24sqrt x}{sqrt x -2}$

$sf ~lim_{xto 4}~~ dfrac{(x -4)(6sqrt x) }{sqrt x - 2}$

$sf ~lim_{xto 4}~~ dfrac{(sqrt x -2)(sqrt x + 2)(6sqrt x) }{sqrt x - 2}$

$sf ~lim_{xto 4}~~(sqrt x + 2)(6sqrt x)$

x= 4 , limit = (√4 + 2)(6√4) = (2 +2)(12) = 48