Question #ff93d

2 Answers
mason m
May 12, 2017

#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#

P dilip_k
May 12, 2017

#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#

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