Question

How do you find the limit `(4x)/(sqrt(2x^2+1))` as `x->oo`?

Answer 1

We have that

`(4x)/(sqrt(2x^2+1))=(4x)/(sqrt(x^2(2+1/x^2)))=(4x)/(absx*(sqrt(2+1/x^2)))`

If x goes to `+oo` the absolute value becomes positive hence `absx=x`
hence

`lim_(x->+oo) (4x)/(absx*(sqrt(2+1/x^2)))= lim_(x->+oo) (4x)/(x*(sqrt(2+1/x^2)))=4/sqrt2=2*sqrt2`

If x goes to `-oo` the absolute value becomes negative hence `absx=-x`
hence

`lim_(x->+oo) (4x)/(absx*(sqrt(2+1/x^2)))= lim_(x->+oo) (4x)/(-x*(sqrt(2+1/x^2)))=-4/sqrt2=-2*sqrt2`