## jpsmarinho 3 years ago 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

2. jpsmarinho

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

3. jpsmarinho

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

4. shubhamsrg

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 ..

5. shubhamsrg

if you understand what am saying.

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

8. jpsmarinho

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

squaring both sides

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@jpsmarinho Now remove the square y and get the radical over whole function .

12. jpsmarinho

Aah right!!

13. jpsmarinho

Other doubt: why I can not consider power the numerator and denominator without equal the expression to some variable?

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