- #1

- 160

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter CrossFit415
- Start date

- #1

- 160

- 0

- #2

VietDao29

Homework Helper

- 1,423

- 2

.... 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.

Your numerator after multiplying by [tex]3 + \sqrt{t}[/tex] is

Oh, and by the way, you should learn https://www.physicsforums.com/showthread.php?t=8997" to make your post much clearer. Just take a look at the page I gave. Learning LaTeX isn't very hard, but I assure that it'll give your post a whole new look. Just give it a try. :)

Last edited by a moderator:

- #3

Mark44

Mentor

- 34,907

- 6,650

.... Lim t approaches 9 9-t/3-sqrt of t.

At the very least, use enough parentheses to make it clear what you mean. Knowledgeable people would reasonably interpret the expression you wrote as

[tex]9 - \frac{t}{3} - \sqrt{t}[/tex]

Even without using LaTeX you could write your limit like this: lim(t -> 9) (9 - t)/(3 - sqrt(t))

- #4

- 160

- 0

Lim t ---> 9

[tex]\frac{(9-t)}{(3-\sqrt{t})}[/tex]

[tex]\frac{(9-t)}{(3-\sqrt{t})}[/tex]

- #5

- 160

- 0

[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.

- #6

eumyang

Homework Helper

- 1,347

- 10

No, don't do that. Instead, notice that when you did multiply the denominator by its conjugate you got

[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.

[tex](3 - \sqrt{t})(3 + \sqrt{t}) = 9 - t[/tex]

?

What you should do is FACTOR the numerator, which is (sort of) a difference of two squares:

[tex]\displaystyle\lim_{t \rightarrow 9} \frac{9 - t}{3 - \sqrt{t}} = \displaystyle\lim_{t \rightarrow 9} \frac{(3 - \sqrt{t})(3 + \sqrt{t})}{3 - \sqrt{t}}[/tex]

Can you take it from here?

Last edited:

- #7

- 3,610

- 1,398

- #8

- 160

- 0

Thanks a lot for your help everyone. It makes a lot of sense now.

Last edited:

- #9

- 160

- 0

Yea I got it

[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

=3+[tex]\sqrt{t}[/tex]

=6

Thank you. :)

[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

=3+[tex]\sqrt{t}[/tex]

=6

Thank you. :)

- #10

Mark44

Mentor

- 34,907

- 6,650

Yea I got it

[tex]\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

= 3+[tex]\sqrt{t}[/tex]

= 6

Thank you. :)

The last two expressions above aren't equal. What you are missing is any indication that you are taking a limit.

What you have above should be written like this.

[tex]\lim_{t \to 9}\frac{(9-t)(3+\sqrt{t}}{(9-t)}[/tex]

[tex]=\lim_{t \to 9}3+\sqrt{t}[/tex]

= 6

If you omit the limit symbols, your instructor is likely to deduct points from your work.

Share: