# Find the limit: `lim_(h->0) (sqrt(81+h)-9)/h`

justaguide | College Teacher | (Level 2) Distinguished Educator

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The limit `lim_(h->0) (sqrt(81+h)-9)/h` has to be determined. If 0 is substituted for h, the indeterminate form 0/0 is obtained. This allows the use of l'Hopital's theorem and the numerator and denominator can be replaced with their derivative.

=> `lim_(h->0) (1/2)/sqrt(81+h)`

substitute h = 0

=> `(1/2)/sqrt 81`

=> `1/18`

The required limit is `1/18`

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The limit `lim_(h->0)(sqrt(81+h) - 9)/h` is required. Substituting h = 0 inĀ  the expression `(sqrt(81+h) - 9)/h` gives `(sqrt(81+0) - 9)/0 = 0/0` which is indeterminate.

We have to change the form of the expression to determine the limit.

`lim_(h->0)(sqrt(81+h) - 9)/h`

= `lim_(h->0)(sqrt(81+h) - 9)/(81 + h - 81)`

= `lim_(h->0)(sqrt(81+h) - 9)/((sqrt(81 + h)^2 - 9^2)`

Use the relation x^2 - y^2 = (x + y)(x - y)

= `lim_(h->0)(sqrt(81+h) - 9)/(((sqrt(81 + h) - 9)(sqrt(81 + h) + 9)))`

= `lim_(h->0)1/(sqrt(81 + h) + 9)`

Substituting h = 0 gives the result `1/(9 + 9) = 1/18`

The required limit `lim_(h->0)(sqrt(81+h) - 9)/h = 1/18`