Solve #lim/(x->0)=(sqrt(9+x) -3)/x#?

Question

How do you solve this limit?

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

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Answer 1 · 2018-03-20T21:44:05

#1/6#

#(sqrt(9+x)-3)/x#

Multiply numerator and denominator by #(sqrt(9+x)+3)#

This is the conjugate of #(sqrt(9+x)-3)#

#((sqrt(9+x)+3)(sqrt(9+x)-3))/(x(sqrt(9+x)+3)#

#x/(x(sqrt(9+x)+3)#

Cancelling:

#cancel(x)/(cancel(x)(sqrt(9+x)+3))=1/(sqrt(9+x)+3)#

Plugging in #x=0#

#1/(sqrt(9+(0))+3)=1/(3+3)=1/6#

#lim_(x->0)((sqrt(9+x)-3)/x)=1/6#

The graph confirms this:

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