How do you evaluate #sqrt(x -1) /( x-1)# as x approaches 1?

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Answer 1 · 2016-07-20T17:05:56

#=lim_(x->1) (sqrt(x - 1))/(x - 1) xx (sqrt(x - 1))/(sqrt(x - 1))#

#=lim_(x->1) (x - 1)/(x - 1 xx sqrt(x - 1))#

#=lim_(x->1)1/(sqrt(x - 1))#

If you look at the limit from both sides, the quantities are unequal, and the function is discontinuous at the point #x = 1#.

Hopefully this helps!

Answer 2 · 2016-07-20T17:42:41

The limit does not exist.

We claim that the limit does not exist.

To see this, observe that #sqrt(x-1)/(x-1)=1/sqrt(x-1)#.

Now, as #xrarr1+, x>1, (x-1)rarr0+, sqrt(x-1)rarr0+#

so that,#1/sqrt(x-1)rarroo#

But, as #xrarr1-,x<1,(x-1)<0, sqrt(x-1)# is not defined.

In view of above, the reqd. limit does not exist.