How do you find the limit of #(1/(3+x)-(1/3))/(x)# as #x->0#?

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Answer 1 · 2016-11-25T20:52:30

#lim_(xrarr0)(1/(3+x)-1/3)/x=-1/9#

#lim_(xrarr0)(1/(3+x)-1/3)/x#

We can clear this limit of fractions by multiplying as follows:

#=lim_(xrarr0)((1/(3+x)-1/3)(3)(3+x))/(x(3)(3+x))#

Multiplying through in the numerator gives:

#=lim_(xrarr0)((3(3+x))/(3+x)-(3(3+x))/3)/(3x(x+3))#

Canceling:

#=lim_(xrarr0)(3-(3+x))/(3x(x+3))#

#=lim_(xrarr0)(3-3-x)/(3x(x+3))#

#=lim_(xrarr0)(-x)/(3x(x+3))#

#=lim_(xrarr0)(-1)/(3(x+3))#

Now we can evaluate the limit, since there is no longer an issue for #xrarr0#.

#=(-1)/(3(0+3))#

#=-1/9#