How do you find `lim sqrt(x^2+1)-x` as `x->oo`?

Let `"f"(x) = (x^2+1)^(1/2) - x`.

Then we can write,

`"f"(x) = (x^2* (1 + 1/x^2))^(1/2)-x`
`"f"(x) = abs(x)*(1+1/x^2)^(1/2)-x`.

Then, we can expand `(1+1/x^2)^(1/2)` using the binomial theorem for `abs(1/x^2)<1`.

`abs(1/x^2)<1` for `1/x^2<1` as `1/x^2` is always positive.
`1/x^2<1` for `x^2>1`, so `x>1` or `x<-1`, i.e. `abs(x) > 1`.

Then for `abs(x)>1`,

`"f"(x) = abs(x) * (1+(1/2)*1/x^2+(1/2)(-1/2)*1/x^4+...) - x`
`"f"(x) = abs(x)-x + 1/2*abs(x)/x^2+(1/2)(-1/2)*abs(x)/x^4+...`

Clearly `lim_{x\to\infty}abs(x)/x^n = 0` for `n>1`.

Then, as this is expansion is valid for any `x>1`,

`lim_{x\to\infty} "f"(x) = abs(x)-x`.

`abs(x)={ (x, x>=0), (-x, x<=0) :}`.

Then, we conclude,

`lim_{x\to+\infty} "f"(x) = 0`,
`lim_{x\to-\infty} "f"(x) = +\infty`.

`lim_(xrarroo)(sqrt(x^2+1)-x)` takes indeterminate form `oo-oo`

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

`= (x^2+1-x^2)/(sqrt(x^2+1)+x)`

`= 1/(sqrt(x^2+1)+x)`

Now use
`sqrt(x^2+1) = sqrt(x^2)sqrt(1+1/x^2)` for all `x != 0`

`= abs(x)sqrt(1+1/x^2)` for all `x != 0`

`= xsqrt(1+1/x^2)` for all `x > 0` (We want `lim_(xrarroo)`)

Returning to

`(sqrt(x^2+1)-x) = 1/(sqrt(x^2+1)+x)`

`= 1/(xsqrt(1+1/x^2))`

As `xrarroo`, we get a limit of `0`

Bonus answer

Ar `xrarr-oo`, the expression `sqrt(x^2+1)-x` takes non-indeterminate form `oo-(-oo) = oo+oo=oo`