Find the limit as x approaches 2 of sqroot(x-2)/(x^2-4)

Expert Answers
jeew-m eNotes educator| Certified Educator

`lim_(xrarr2) sqrt((x-2)/(x^2-4))`

`=lim_(xrarr2) sqrt((x-2)/(x^2-2^2))`

`=lim_(xrarr2) sqrt(((x-2))/((x+2)(x-2)))`

`=lim_(xrarr2) sqrt(1/(x+2))`

`= sqrt(1/(2+2))`

`= sqrt(1/4)`

`= 1/2`

 

Therefore;

`lim_(xrarr2) (sqrt((x-2)/(x^2-4))) = 1/2`

nick-teal eNotes educator| Certified Educator

This limit does not exist.  The limit from the right exists, but the left does not.  There are no values for this function for x<2.

I think the square root was supposed to cover the entire problem.

tonys538 | Student

The limit `lim_(x->2) sqrt(x-2)/(x^2-4)` has to be determined.

If we substitute x = 2, we get the result `sqrt(2-2)/(2^2 - 4) = 0/0` .

0/0 is an indeterminate form that allows us to use l'Hopital's rule and substitute the numerator and denominator by their derivatives.

This gives:

`lim_(x->2) ((1/2)*(1/sqrt(x-2)))/(2x)`

`lim_(x->2) 1/(4*x*sqrt(x-2))`

Substituting x = 2 gives `1/0 = oo`

The limit `lim_(x->2) sqrt(x-2)/(x^2-4) = oo`