`lim_(x->0+)(sqrt(10+h)-sqrt10)/sqrt(h)`

The symal 0+ is the right-hand limit.

Give your answer to 2 decimal place

### 2 Answers | Add Yours

Multiply by the conjugate of the top:

[sqrt(10+h)-sqrt(10)] * [sqrt(10+h)+sqrt(10)]/

[sqrt(h) * sqrt(10+h)+sqrt(10)]

This gives:

(10 + h - 10)/(sqrt(h) * sqrt(10+h)+sqrt(10))

h/(sqrt(h) * sqrt(10+h)+sqrt(10))

sqrt(h) / sqrt(10+h)+sqrt(10))

When we take the limit at 0, we get:

sqrt 0 / sqrt 10 + sqrt 10

0/(2sqrt 10) = 0

So, this limit = 0

`lim_(h->0+)(sqrt(10+h)-sqrt(10))/sqrt(h)`

`=lim_(h->0+)((sqrt(10+h)-sqrt(10))(sqrt(10+h)+sqrt(10)))/(sqrt(h)(sqrt(10+h)+sqrt(10)))`

`=lim_(h->0+)(10+h-10)/(sqrt(h)(sqrt(10+h)+sqrt(10)))`

`=lim_(h->0+)sqrt(h)/(sqrt(10+h)+sqrt(10))`

`=0/(sqrt(10)+sqrt(10))`

`=0`

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