Q. `lim_(x->oo)` (`sqrt(x+1)` - ```sqrt(x)` )



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tiburtius's profile pic

Posted on (Answer #1)

If you have limit `lim (f-g)=oo-oo` then you can rewrite this as product `lim f[1-g/f]` now you can solve `lim g/f` by using L'Hospital's rule (`lim (f/g) =lim (f'/g')` ).

`lim_(x->oo)(sqrt(x+1)-sqrt(x))=lim_(x->oo)sqrt(x+1)(1-(sqrtx)/(sqrt(x+1)))`` `

Now we solve `lim_(x->oo)(sqrtx)/(sqrt(x+1))=1` (we get this by dividing both numerator and denominator by `sqrtx` )

Now our limit becomes `oo cdot 0` so we need to rewrite it again.


Now we have limit of form `0/0` so we can use LHospital's rule.

`lim_(x->oo)(sqrt[x]/(2 (1 + x)^(3/2)) - 1/(2 sqrt[x] sqrt[1 + x]))/(-(1/(2 (1 + x)^(3/2))))=lim_(x->oo)1/sqrtx=0`

Hence, your result is:  `lim_(x->oo)(sqrt(x+1)-sqrt(x))=0`

Top Answer

aruv's profile pic

Posted on (Answer #2)

Let us write


`lim x-> oo ==> y->0`




`` `=lim_(y->0){(sqrt(y+1)-1)(sqrt(y+1)+1)}/{sqrt(y)(sqrt(y+1)+1)}`





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