How do you find the limit of ((1/x+16)-(1/16)) / x(1x+16)(116)x as x approaches 0?

1 Answer
May 22, 2015

I think you meant: (1/(x+16)-1/16) / x1x+16116x.

When we try to find the limit, we get the (indeterminate) form 0/000. 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)1x+16116x=1616(x+16)x+1616(x+16)x1

= ((16- (x+16))/(16(x+16)))/(x/1)=16(x+16)16(x+16)x1

= (-x)/(16(x+16))*1/x=x16(x+16)1x

= (-1)/(16(x+16))=116(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)16(x+16), so:

(1/(x+16) - 1/16)/x = ((1/(x+16) - 1/16))/x (16(x+16))/(16(x+16)1x+16116x=(1x+16116)x16(x+16)16(x+16)

= ((16(x+16))/(x+16) - (16(x+16))/16)/(x(16(x+16))=16(x+16)x+1616(x+16)16x(16(x+16))

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

= (-1)/(16(x+16))=116(x+16)

So, we get:

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