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# `lim_(x->0+)(sqrt(10+h)-sqrt10)/sqrt(h)` The symal 0+ is the right-hand limit. Give...

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

The symal 0+ is the right-hand limit.

Posted by cck123 on October 27, 2013 at 3:03 AM via web and tagged with limit, math

College Teacher

(Level 2) Assistant Educator

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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

Posted by steveschoen on October 27, 2013 at 4:52 AM (Answer #2)

High School Teacher

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`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`

``

Posted by aruv on October 27, 2013 at 5:55 AM (Answer #3)