Question

Evaluate the limit, if it exists. (if an answer does not exist, enter dne.) lim h→0 (5 + h)−1 − 5−1 h

Answer 1

`\displaystyle\lim_(h\to0)\frac{(5+h)^(-1)-5^(-1)}h=\lim_(h\to0)\frac{\frac5{5(5+h)}-(5+h)/(5(5+h))}h`

`\displaystyle=-\lim_(h\to0)\frac h{5(5+h)h}`

`\displaystyle=-\lim_(h\to0)\frac1{5(5+h)}=-\frac1{25}`

Alternatively, recall that if
`f(x)=\frac1x`, then
`f'(x)=-\frac1{x^2}`, and so


`f'(5)=\displaystyle\lim_(x\to5)(\frac1x-\frac15)/(x-5)`

Take
`h=x-5`, so that
`x=h+5`, and we have the original limit. So the limit is equivalent to the value of
`f'(5)`, i.e.


`f'(5)=-\frac1{5^2}=-\frac1{25}`