Question

Limits - check my work

Answer 1

\[\lim \frac{ \sqrt(6-x)-2 }{ \sqrt(3-x)-1 }\]
As x approaches 2.

Answer 2

Is the limit zero?

Answer 3

Also, the limit as x approaches 2 from the left of:
\[\frac{ x^2-2x }{ x^2-4x+4 }\]
The limit is -infinity, right?

Answer 4

If you want to put more than one character under a root, then you have to use curly braces.

Example:
\sqrt(6-x) = \(\sqrt(6-x)\)
but
\sqrt{6-x} = \(\sqrt{6-x}\)

Answer 5

\[\large\rm \lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\]If you plug in the value 2,
the limit is approaching the `indeterminate form` 0/0.

0/0 is not a number. It is certainly not 0.
So we try some algebra.
Hmm thinking.

Answer 6

Hmm ok this one was not as obvious as I thought it was going to be.
You'll want to multiply both numerator and denominator by the `conjugate` of both the numerator and denominator.\[\large\rm \lim_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\left(\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\right)\left(\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right)\]Only multiply out the ones that are conjugates.
Don't expand out the messy stuff like this \(\rm (\sqrt{6-x}-2)(\sqrt{3-x}+1)\)
Just leave those sitting next to each other and deal with the conjugates.

Answer 7

For your other problem,\[\large\rm \frac{x^2-2x}{x^2-4x+4}=\frac{x\cancel{(x-2)}}{\cancel{(x-2)}(x-2)}\]

Answer 8

We're approaching 2/0 so this one is blowing up infinitely large.
And yes, you have the right idea,
since we're approaching from the left and our x is slightly smaller than 2,
the denominator ends up giving us a negative.

Answer 9

Thanks zep :)
So for the first problem, multiplying by the conjugates is what I originally tried, but I must have made a mistake... here goes again.

Answer 10

Wait, hold up. Multiply by the conjugate of the numerator AND denominator?

Answer 11

you could also try using l'hopitals rule for the first one.

Answer 12

What is thatZ o.O

Answer 13

This is way early for L'Hopital.
She's not that far into calc yet D:

Yes conjugate of numerator AND denominator as I showed.

Answer 14

Haha. Zep is right - I'm a newbie :3

Answer 15

Look good so far? (sorry for the bad formatting - I'm on the bus right now)
(2-x)(sqrt(3-x) + 1) over (2 - x)(sqrt(6-x) + 2
And then cancel out the (2-x)...
(sqrt(3+x) + 1) over (sqrt(6-x) + 2)
Right?

Answer 16

Yesss

Answer 17

Then plug in 2?
:D

Answer 18

Yesss \c:/

Answer 19

Gotta run! Thanks zebbles
:)

Answer 20

Interwebs....... on the bus? 0_o
What is this madness?

Answer 21

1/2 zep?

Answer 22

yay good job

Answer 23

Haha. I have to utilize my bus time xD desperate times call for desperate measures

Answer 24

hah :D

Answer 25

And the second one was -inf, right?