Question

How do you find `lim t(sqrt(t+1)-sqrtt)` as `t->oo`?

Answer 1

`lim_(t->oo) t(sqrt(t+1)-sqrt(t)) = oo`

Explanation:

`lim_(t->oo) t(sqrt(t+1)-sqrt(t))`

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

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

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

`=lim_(t->oo) sqrt(t) * sqrt(t)/(sqrt(t+1)+sqrt(t))`

`=lim_(t->oo) sqrt(t) * 1/(sqrt(1+1/t)+1)`

`=lim_(t->oo) 1/2sqrt(t)`

`=oo`

Answer 2

`+oo`

Explanation:

I'll replace `t` with `x`

`lim_(xrarr+oo)x*(sqrt(x+1)-sqrtx)` `=`

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

`lim_(xrarr+oo)x*(sqrt(x+1)^2-sqrtx^2)/(sqrt(x+1)+sqrtx)` `=`

`lim_(xrarr+oo)x*(cancel(x)+1-cancel(x))/(sqrt(x+1)+sqrtx)` `=`

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

`lim_(xrarr+oo)x/(sqrt(x^2(1/x+1/x^2))+sqrt(x^2*(1/x)))` `=`

`lim_(xrarr+oo)x/(|x|sqrt(1/x+1/x^2)+|x|sqrt(1/x))`

`x->+oo` , `x>0`

`=` `lim_(xrarr+oo)x/(xsqrt(1/x+1/x^2)+xsqrt(1/x))` `=`

`lim_(xrarr+oo)cancel(x)/(cancel(x)(sqrt(1/x+1/x^2)+sqrt(1/x))` `=`

`lim_(xrarr+oo)1/(sqrt(1/x+1/x^2)+sqrt(1/x))` `=` `+oo`

because `lim_(xrarr+oo)x=+oo` so `lim_(xrarr+oo)1/x=0`

and `lim_(xrarr+oo)sqrt(1/x)=0`