Find the limit as x approaches 3 from the right #ln(x^2-9)#?

Question

I keep on getting oo, however the answer is -oo. Isnt #(x^2-9)# going to give a postive

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Answer 1 · 2017-03-24T23:48:36

Please see below.

For #u > 1#, we get #lnu >0#, but for #0 < u < 1#, we have #ln u < 0#.

So #lim_(urarr0^+) lnu = -oo#

As #xrarr3^+#, we have #(x^2 - 9) rarr 0^+#

Therefore, #lim_(xrarr3^+)ln(x^2-9) = -oo#

A little more to think about

For #x = sqrt10 ~~ 3.1623#, we have #x^2 -9 = 1#, so #ln(x^2 - 9) = ln1 = 0#

As #x# continues toward #3# from the right, we have #x^2# is between #9# and #10#, so #(x^2 - 9)# is between #0# and #1# and its ln is negative.