Question

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

Answer 1

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`.