`lim_(h -> 0) (sqrt(9 + h) - 3)/h` Evaluate the limit, if it exists.

Textbook Question

Chapter 2, 2.3 - Problem 21 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

`lim_(h -> 0) (sqrt(9 + h) - 3)/h`

sol:

`lim_(h -> 0) (sqrt(9 + h) - 3)/h`

considering the numerator, we get

`(sqrt(9 + h) - 3)`

= `(sqrt(9 + h) - 3) *(((sqrt(9 + h) + 3))/((sqrt(9 + h) + 3)))`

= `((sqrt(9 + h))^2 - (3)^2)/((sqrt(9 + h) + 3))`

= `h/((sqrt(9 + h) + 3))`

so,

`lim_(h -> 0) (sqrt(9 + h) - 3)/h`

= `lim_(h -> 0) (h/((sqrt(9 + h) + 3)))/h`

= `lim_(h -> 0) 1/((sqrt(9 + h) + 3))`

now as `h-> 0` we get ,

`lim_(h -> 0) 1/((sqrt(9 + h) + 3))`

=  `1/((sqrt(9 + 0) + 3)) = 1/(3+3) =1/6`

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

Plugging in h = 0 into the function results in the indeterminate form `0/0` . Since the numerator and denominator are both differentiable functions, we can apply L'Hopital's rule. 

`lim_(h->0) ( (1/2)(9 + h)^(-1/2) )/ (1)` is equivalent to the original problem. Plugging in h = 0 shows that the limit is `1/6` . 

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