Solve Limit Problem: 9-t/3-sqrt(t) & x^2-81/sqrt(x)-3 | Expert Tips & Tricks

[ocean09]

hi guys,

I need help w/ finding the limit for the following problem:

lim 9-t/3- radical t =
t-->9

lim x^2-81/radical x -3
t-->0
Another question, how do you guys do those

 

[LeonhardEuler]

I'm not sure I understand the question. Is the first one:
$$\lim_{t\rightarrow 9} 9-\frac{t}{3} -\sqrt{t}$$?
If so, this is not difficult because it is the sum of three continuous functions.
For the second, is it:
$$\lim_{x\rightarrow 0} x^2-\frac{81}{\sqrt{x}}-3$$?
If so, then this is also not difficult since the limits of the first and last terms are finite while that of the middle term is not. (I'm assuming that 't' in the second one is supposed to be an 'x'. If not then I don't understand.)

To see how to make these math symbols, click on them.

 

[Data]

probably safer to assume the limits he wants are

$$\lim_{t \rightarrow 9} \frac{9-t}{3-\sqrt{t}}$$

and

$$\lim_{x \rightarrow 0} \frac{x^2-81}{\sqrt{x}-3}.$$

The second one is continuous at 0 so you can just sub x=0 in. Are you sure it's not $x \rightarrow 9$ again?

For the first one, you can factor it. I'll let you try for yourself first.

 

[ocean09]

probably safer to assume the limits he wants are

$$\lim_{t \rightarrow 9} \frac{9-t}{3-\sqrt{t}}$$

and

$$\lim_{x \rightarrow 0} \frac{x^2-81}{\sqrt{x}-3}.$$

You are correct.

The second one is continuous at 0 so you can just sub x=0 in. Are you sure it's not $x \rightarrow 9$ again?

oops. That was a typo. it's x-->9

For the first one, you can factor it. I'll let you try for yourself first.

Factor? I thought I'm supposed to multiply top and bottom by radical x -3...correct me if I'm wrong...



This might be a stupid question.

How do you guys show your work like that? Is there a special program?

Thanks.