How do you find the limit of #((1/x+16)-(1/16)) / x# as x approaches 0?

1 Answer
Jim H
May 22, 2015

I think you meant: #(1/(x+16)-1/16) / x#.

When we try to find the limit, we get the (indeterminate) form #0/0#. We need to rewrite and try again.

The important algebra can be done in a couple ways:

Method 1

"If I had a fraction over a fraction, I'd know what to do next."

Good! Make it so.

#(1/(x+16) - 1/16)/x = (16/(16(x+16)) - (x+16)/(16(x+16)))/(x/1)#

#= ((16- (x+16))/(16(x+16)))/(x/1)#

#= (-x)/(16(x+16))*1/x#

#= (-1)/(16(x+16))#

Method 2

"I know this trick:"
Multiply numerator and denominator by the common denominator of all the fractions in the numerator and denominator. (Sounds complicated, but it's quicker.)

The common denominator is #16(x+16)#, so:

#(1/(x+16) - 1/16)/x = ((1/(x+16) - 1/16))/x (16(x+16))/(16(x+16)#

#= ((16(x+16))/(x+16) - (16(x+16))/16)/(x(16(x+16))#

#=(16-(x+16))/(16x(x+16))#

#= (-1)/(16(x+16))#

So, we get:

#lim_(xrarr0) (1/(x+16)-1/16) / x = lim_(xrarr0) (-1)/(16(x+16)) = (-1)/16^2 = (-1)/256#.

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