- #1
- 160
- 0
.... Lim t approaches 9 9-t/3-sqrt of t. First I multiplied the conjugate of 3 + sqrt of t to both numerator and denominator. I got 27+9 sqrt of t - 4t / 9-t .... I plug 9 to t. Denominator cancels out. I got 18 but answer states 6. How?
Your numerator after multiplying by [tex]3 + \sqrt{t}[/tex] is incorrect!! You shouldn't expand everything out like that. After multiplying, keep the numerator, and try canceling out instead..... Lim t approaches 9 9-t/3-sqrt of t. First I multiplied the conjugate of 3 + sqrt of t to both numerator and denominator. I got 27+9 sqrt of t - 4t / 9-t .... I plug 9 to t. Denominator cancels out.
At the very least, use enough parentheses to make it clear what you mean. Knowledgeable people would reasonably interpret the expression you wrote as.... Lim t approaches 9 9-t/3-sqrt of t.
No, don't do that. Instead, notice that when you did multiply the denominator by its conjugate you gotHere's how I did it...
[tex]\frac{9-t}{3-\sqrt{t}}[/tex] ([tex]\frac{3+\sqrt{t}}{3+\sqrt{t}}[/tex]) =
[tex]\frac{27+9\sqrt{t}-3t-t\sqrt{t}}{9-t}[/tex]
Plugged in 9 for t...
then I got a different answer than 6.
The last two expressions above aren't equal. What you are missing is any indication that you are taking a limit.Yea I got it
[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]
= 3+[tex]\sqrt{t}[/tex]
= 6
Thank you. :)