Question

How do you find the limit of `sqrt(2-x^2)/x` as `x->0^+`?

Answer 1

`lim_(x->0^+) sqrt(2-x^2)/x = +oo`

Explanation:

For `x > 0`:

`sqrt(2-x^2)/x > 0`

Besides the numerator is bounded in the interval `(0,2)` while the denominator is infinitesimal, so the quotient is positive and unbounded and therefore its limit is infinite.

Thus for any `M in (0,+oo)`, if we choose `delta_M` such that:

`delta_M < sqrt(2/(1+M^2))`

then we can see that:

`delta_M^2 (1+M^2) < 2`

`M < sqrt(2-delta_M^2)/delta_M`

Then for `x in (0, delta_M)`

`sqrt(2-x^2)/x > sqrt(2-delta_M^2)/delta_M > M`

Which proves that:

`lim_(x->0^+) sqrt(2-x^2)/x = +oo`

graph{sqrt(2-x^2)/x [-10, 10, -5, 5]}