Question

Question #ff93d

Answer 1

`4`

Explanation:

Multiply the numerator and denominator by the conjugate of the denominator:

`=lim_(xrarr2)((x-sqrt(8-x^2))(sqrt(x^2+12)+4))/((sqrt(x^2+12)-4)(sqrt(x^2+12)+4))`

`=lim_(xrarr2)((x-sqrt(8-x^2))(sqrt(x^2+12)+4))/((x^2+12)-16)`

`=lim_(xrarr2)((x-sqrt(8-x^2))(sqrt(x^2+12)+4))/(x^2-4)`

The denominator is still `0` as `xrarr2`, so let's also try multiplying by the conjugate of the original numerator:

`=lim_(xrarr2)((x-sqrt(8-x^2))(x+sqrt(8-x^2))(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))`

`=lim_(xrarr2)((x^2-(8-x^2))(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))`

`=lim_(xrarr2)((2x^2-8)(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))`

`=lim_(xrarr2)(2(x^2-4)(sqrt(x^2+12)+4))/((x^2-4)(x+sqrt(8-x^2))`

`=lim_(xrarr2)(2(sqrt(x^2+12)+4))/(x+sqrt(8-x^2)`

Now we can evaluate the limit:

`=(2(sqrt(4+12)+4))/(2+sqrt(8-4))`

`=(2(4+4))/(2+2)`

`=4`

Answer 2

`lim_(xto2) (x-sqrt(8-x^2))/(sqrt(x^2+12)-4)`

`=lim_(xto2) (x-sqrt(8-x^2))/(sqrt(x^2+12)-4)xx (x+sqrt(8-x^2))/(sqrt(x^2+12)+4)xx(sqrt(x^2+12)+4)/(x+sqrt(8-x^2))`

`=lim_(xto2) (x^2-8+x^2)/(x^2+12-4^2)xx(sqrt(x^2+12)+4)/(x+sqrt(8-x^2))`

`=lim_(xto2) (2(cancel(x^2-4)))/((cancel(x^2 -4)))xx(sqrt(x^2+12)+4)/(x+sqrt(8-x^2))`

[`" "as" " xto2=>x!=2=>x^2-4!=0` ]

so the limit becomes

`=lim_(xto2) (2(sqrt(x^2+12)+4))/(x+sqrt(8-x^2))`

`= (2(sqrt(2^2+12)+4))/(2+sqrt(8-2^2))`

`= (2(sqrt16+4))/(2+sqrt(8-4))`

`=(2xx8)/(2+2)=4`