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

Answer 1

#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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Answer 2

Elijah Abram

To find the limit of x/sqrt(9-x^2) as x approaches -3+, we can substitute the value -3 into the expression and simplify it. By doing so, we get (-3)/sqrt(9-(-3)^2), which simplifies to -3/sqrt(9-9), and further simplifies to -3/0. Since we have a denominator of 0, this indicates that the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.