How do you find #lim (sqrt(9-x)-3)/x# as #x->0# using l'Hospital's Rule?

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Answer 1 · 2016-12-25T06:58:54

The answer is #=-1/6#

We need

#(sqrtx)'=1/(2sqrtx)#

#lim_(x->0)(sqrt(9-x)-3)/x=0/0#

This is undefined, so we apply L'Hôspital's Rule

#lim_(x->0)(sqrt(9-x)-3)/x=lim_(x->0)((sqrt(9-x)-3)')/(x')#

#=lim_(x->0)(-1/(2sqrt(9-x)-3))#

#=-1/6#