How do you find the #lim_(x to 1) 1/(sqrt (x-3) + sqrt 3)#?

Question

736 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-12T14:30:59

You merely evaluate the expression at #x = 1#; this will produce a complex number of the form #1/(a+bi)#.

Given: #lim_(x to 1) 1/(sqrt (x-3) + sqrt 3)#

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt (1-3) + sqrt 3)#

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt (-2) + sqrt 3)#

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt3+ sqrt2i)#

Multiply by 1 in the form of #(sqrt3- sqrt2i)/(sqrt3- sqrt2i)#:

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = 1/(sqrt3+ sqrt2i)(sqrt3- sqrt2i)/(sqrt3- sqrt2i)#

We do this because we know that multiplication by the conjugate converts the denominator into a real number.

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = (sqrt3- sqrt2i)/(3- 2)#

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = (sqrt3- sqrt2i)/1#

#lim_(x to 1) 1/(sqrt (x-3) + sqrt 3) = sqrt3- sqrt2i#