How do you find the limit of # ((1/y) - (1/7))/(y-7)# as y approaches 7?

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Answer 1 · 2016-02-04T20:34:58

#-1/49#

This can be expressed as

#lim_(yrarr7)(1/y-1/7)/(y-7)#

We should try to clear the fractions from the denominator by multiplying the fraction by #(7y)/(7y)#.

#=lim_(yrarr7)((7y)/y-(7y)/7)/(7y(y-7))#

This simplifies to be

#=lim_(yrarr7)(7-y)/(7y(y-7))#

Note that this can be simplified, since #7-y=-(y-7)#.

#=lim_(yrarr7)(-(y-7))/(7y(y-7))#

The #y-7# terms will cancel.

#=lim_(yrarr7)(-1)/(7y)#

The limit can now be evaluated by plugging in #7# for #y#.

#=(-1)/(7(7))#

#=-1/49#

We can check the graph of the function. Even though the point at #7# is undefined, it should be at #-1/49# #(#very close to #0)#.

graph{(1/x-1/7)/(x-7) [6.347, 7.6793, -0.3486, 0.317]}