Q. `lim_(x->oo)` (`sqrt(x+1)` - ```sqrt(x)` )
2 Answers
tiburtius | High School Teacher | (Level 2) Educator
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.
`lim_(x->oo)(1-(sqrtx)/(sqrt(x+1)))/(1/(sqrt(x+1)))=`
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`
Let us write
`x=1/y`
`lim x-> oo ==> y->0`
`Thus`
`lim_(x->oo)(sqrt(x+1)-sqrt(x))=lim_(y->0)(sqrt(1+1/y)-sqrt(1/y))`
`=lim_(y->0)(sqrt(y+1)-1)/sqrt(y)`
`` `=lim_(y->0){(sqrt(y+1)-1)(sqrt(y+1)+1)}/{sqrt(y)(sqrt(y+1)+1)}`
`=lim_(y->0)y/{sqrt(y)(sqrt(y+1)+1)}`
`=lim_(y->oo)sqrt(y)/(sqrt(y+1)+1)`
`=0`
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