1 answer

How do you find the limit of #x/sqrt(9-x^2)# as x approaches -3+?

Question:

How do you find the limit of #x/sqrt(9-x^2)# as x approaches -3+?

Answers

#hlim_(xrarr-3^+) x/sqrt(9-x^2) = -oo# (Does not exist because as xrarr-3^+ , the ratio decreases without bound.)

As #xrarr-3^+#, the numerator approaches #-3# and the denominator approaches #0# through positive values. So the fraction is negative and getting more negative as #xrarr-3#.

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