## gabie1121 2 years ago find the limit as x approaches 9 : WILL GIVE EQUATION IN RESPONSE

1. gabie1121

$\lim_{x \rightarrow 9}(\frac{ 1 }{ \sqrt{x}-3 } - \frac{ 6 }{ x-9 })$

2. AravindG

do you know l hospital's rule?

3. gabie1121

please don't use lhopitals or anything im in basic calculus and haven't learned it yet. Suposed to be able to algebriaically change it to solve. I know the answer is suposed to be 1/6 i just don't know how to get there

4. AravindG

ok by basics , rationalise first fraction

5. AravindG

multiply by $$\sqrt{x}+3$$ on numerator and denominator

6. gabie1121

so the left turns into right?

7. gabie1121

ugh didn't get the equation in

8. gabie1121

$\frac{ \sqrt{x}+3 }{ x-3 }$

9. AravindG

try again ?

10. gabie1121

is that not right after multiplying by $\sqrt{x}$+3?

11. gabie1121

so i now have $\frac{ \sqrt{x}+3 }{ x-9 } - \frac{ 6 }{ x-9}$

12. AravindG

ok now add them !! note denominator is same

13. gabie1121

well that gives me zero... and i know the answer is suposed to be 1/6

14. AravindG

try again?

15. gabie1121

right, nevermind. thanks!

16. gabie1121

no i still get zero

17. gabie1121

$\frac{ \sqrt{x}+3 }{ x-9 }- \frac{ 6 }{ x-9 }$

18. AravindG

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19. gabie1121

$\frac{ \sqrt{x}-3 }{ x-9 }$ and if i plug in 9 it will be 0 over 0

20. AravindG

please dont jump into putting vale ! we are not over yet!

21. AravindG

*value

22. gabie1121

ok, what do i do if i dont put the value in then

23. gabie1121

because your calculations give me $\lim_{x \rightarrow 9}\frac{ \sqrt{x}-3 }{ x-9 }$

24. gabie1121

do i split the bottom up? into its factors?

25. gabie1121

found it out! thanks!!

26. AravindG

yw : )

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