How do you find #lim (sqrtx-1)/(x-1)# as #x->1^+# using l'Hospital's Rule or otherwise?

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Answer 1 · 2016-12-09T14:33:17

I wouldn't use l'Hospital for this.

I would factor: #x-1 = (sqrtx+1)(sqrtx -1)#

#lim_(xrarr1^+) (sqrtx-1)/(x-1) = lim_(xrarr1) (sqrtx-1)/((sqrtx-1)sqrtx+1)#

# = lim_(xrarr1) 1/(sqrtx+1) = 1/(1+1) = 1/2#

(Of course, I wouldn't write all that if I wasn't trying to explain.

#lim_(xrarr1^+)(sqrtx-1)/(x-1) = lim_(xrarr1)1/(sqrtx+1) = 1/2#

OR

"Rationalize the numerator by multiplying by #(sqrtx+1)/(sqrtx+1)# to get

#lim_(xrarr1^+) (sqrtx-1)/(x-1) = lim_(xrarr1) overbrace(cancel(x-1))^1/(cancel((x-1)) (sqrtx+1)) = 1/2#