Question

HELPP!!

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.
Match the resulting values to the corresponding limits.

Answer 1

Answer:
`\lim_(x \to \infty) (4x-1)/(9+12x)\to (1)/(3)`

`\lim_(x \to 4) (x^2-x-12)/(5x-20)\to(7)/(5)`

`\lim_(x \to -\infty) (5x+6)/(4x+2)\to(5)/(4)`

`\lim_(x \to 3) (4x-12)/(x^2-9)\to(2)/(3)`

Explanation:

i)
`\lim_(x \to \infty) (4x-1)/(9+12x)`

Taking x as common outside from numerator and denominator , we have

`\lim_(x \to \infty) (x(4-(1)/(x)))/(x((9)/(x)+12))\\\\=\lim_(x \to \infty) (4-(1)/(x))/((9)/(x)+12)\\\\=(4-0)/(0+12)=(4)/(12)=(1)/(3)`

ii)
`\lim_(x \to 4) (x^2-x-12)/(5x-20)`

It can be written as :
`\lim_(x \to 4) ((x-4)(x+3))/(5(x-4))`

`=\lim_(x \to 4) (x+3)/(5)=(4+3)/(5)=(7)/(5)`

iii)
`\lim_(x \to -\infty) (5x+6)/(4x+2)`

Taking x as common outside from numerator and denominator , we have

`\lim_(x \to -\infty) (x(5+(6)/(x)))/(x(4+(2)/(x)))\\\\=(5+0)/(4+0)=(5)/(4)`

iv)
`\lim_(x \to 3) (4x-12)/(x^2-9)`

which can be written as :

`\lim_(x \to 3) (4(x-3))/((x-3)(x+3))` ∵
`a^2-b^2=(a+b)(a-b)`

`=\lim_(x \to 3) (4)/((x+3))=(4)/(3+3)=(4)/(6)=(2)/(3)`