What is #lim_(x->4)# of #(x-4)/(sqrtx-2)#?

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Answer 1 · 2016-02-13T18:15:09

#4#

Since this limit is an indeterminate form of type #0/0#, we may use L'Hospital's Rule to evaluate the limit as follows :

#lim_(x->4)(x-4)/(sqrtx-2)=lim_(x->4)(d/dx(x-4))/(d/dx(sqrtx-2))#

#=lim_(x->4)1/(1/2x^(-1/2))#

#=4#.

Answer 2 · 2016-02-13T18:52:07

#4#

If we want to refrain from using calculus, we can determine the limit algebraically by multiplying the function by the conjugate of the denominator.

#=lim_(xrarr4)((x-4)(sqrtx+2))/((sqrtx-2)(sqrtx+2))#

The denominator is now in the form #(a-b)(a+b)=a^2-b^2#.

#=lim_(xrarr4)((x-4)(sqrtx+2))/(x-4)#

The #x-4# terms will cancel.

#=lim_(xrarr4)sqrtx+2#

We can now evaluate the limit by plugging #4# into the equation.

#=sqrt4+2=4#

We can check a graph of the original function:

graph{(x-4)/(sqrtx-2) [-2.875, 17.125, -1.44, 8.56]}

#(4,4)# is undefined, but it is where the function would be.