limit x->2 (sqrt(6-x)-2)/(sqrt(3-x)-1)
Cannot use L-hospital rule or deriviates :/
You rationalize the denominator by mult top and bottom of the fraction
by (conjugate of denominator) [√(3-x) +1].
After simplifying you find the numerator has a limit of 2, as the
denominator approaches zero. Therefore there is no limit.
BTW you use your grouping symbols astutely (you might mix parentheses
with alternatives), but too often questions (and answers) are terribly ambiguous
because of LACK OF GROUPING. Thanks!
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You rationalize the denominator by mult top and bottom of the fraction
by (conjugate of denominator) [√(3-x) +1].
After simplifying you find the numerator has a limit of 2, as the
denominator approaches zero. Therefore there is no limit.
BTW you use your grouping symbols astutely (you might mix parentheses
with alternatives), but too often questions (and answers) are terribly ambiguous
because of LACK OF GROUPING. Thanks!