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

1 Answer
Andrea S.
Jun 1, 2017

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]}

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
1595 views around the world
You can reuse this answer
Creative Commons License