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]}

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