lim sqrt(x^2-9)/(2x-6) when x goes to -infinity. I use a software and discover that the lim equals to -1/2, but, when I do, II only encounter the result 1/2

Question

anonymous · Answer

attachment

jpsmarinho · Answer

Ah but has square root in the x^2 -9

jpsmarinho · Answer

The lim is the following: {(x^2-9)^(1/2)}/(2x-6)

shubhamsrg · Answer

well it should be -1/2 only
cant be 1/2 as for x tending to -(a very large no.) ,,the expression gets negative,,clearly..
you can justify this as you may 4 = (-2)^2 and send 4 under sqrt sign with - aside ..

shubhamsrg · Answer

if you understand what am saying.

anonymous · Answer

For positive infinity it is 1/2 for negative infinity it is -1/2

anonymous · Answer

@jpsmarinho     This is answer of your question..

jpsmarinho · Answer

What you did in first step? I don't understand.

anonymous · Answer

squaring both sides

anonymous · Answer

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