## zned6559 one year ago Calculate lim t-> negative infinity ((2t^2 - t -2)^1/2) / (2t+1)

1. amistre64

x = sqrt(x^2) ... might be useful

2. zned6559

I know the answer is -1/sqrt 2 but I don't know how to get there

3. amistre64

^(1/2) is the same as sqrt() id put them both under the sqrt and compare them

4. amistre64

$\frac{\sqrt{2t^2-t-1}}{\sqrt{(2t+1)^2}}$

5. zned6559

the bottom is suppose to be 2t+1 no sqrt

6. amistre64

what does sqrt(x^2) reduce to?

7. zned6559

x

8. amistre64

then sqrt((2t+1)^2) = 2t+1

9. zned6559

$\sqrt{2t ^{2}-t-2}\div2t+1$

10. amistre64

you are not seeing the obvious ... $\sqrt{2t ^{2}-t-2}\div2t+1$ $\sqrt{2t ^{2}-t-2}\div \sqrt{(2t+1)^2}$

11. amistre64

what is the expansion of (2t+1)^2 ?

12. zned6559

$4t ^{2}+4t+1$

13. amistre64

then we have $\sqrt{\frac{2t^2-t-2}{4t^2+4t+1}}$

14. amistre64

what do we know about polynomials and their limits?

15. zned6559

not sure

16. amistre64

then you need to cover some of your basic rules .... not sure what theyve covered tho

17. amistre64

polys of the same degree, limit to their leading coeffs

18. zned6559

I got it now!

19. amistre64

the only thing to watch for is the sign, by squaring the bottom we altered the sign of it inthe process.

20. amistre64

2t+1 as t to -inf is a negative value .... so we are originally going to have a negative limit in the end