Question

lim as x approaches 3 of (1/x-3)(1/sqrt(x+1)-1/2)

Answer 1

\[\lim_{x \rightarrow 3} \frac{ 1 }{ x-3 }(\frac{ 1 }{ \sqrt{x+1} }-\frac{ 1 }{ 2 })\]

Answer 2

so what type of indeterminant to you get?

Answer 3

to start, so we can see if it is something we can work with :)

Answer 4

What?

Answer 5

I just know that you're supposed to multiply by the conjugate.

Answer 6

Fib is asking if something bad happens when you plugin 3 into the expression

Answer 7

that is what I am asking essentially^ :)

Answer 8

0/0

Answer 9

how are you getting a zero on top? but besides the point

Answer 10

I just put that, but it comes out to just zero.

Answer 11

It's 0/0

Answer 12

When you plug in the 3, you get 1/0 (0) so it's 0/0, isn't it?

Answer 13

I just need help getting passed the first step.

Answer 14

Yes, it is \(\frac 00\).

Answer 15

no, 1/0 is just 1/0.

Answer 16

But times the zero inside of the parenthesis?

Answer 17

yes it is times zero, but that is an inderterminant form, times zero. so we can't just say that as eaily

Answer 18

Isn't this a little beside the point, though?

Answer 19

so personally I would use squeeze theorem here, have you learned it?

Answer 20

\(1/0\) is not an indeterminate form.

Answer 21

And \(1\times 0 = 0\)

Answer 22

1/0 is inderterminant. 0/0 is underfined

Answer 23

crap got em backwards yet again didn't I?

Answer 24

...

Answer 25

ugh. I always do that, but doesn't really matter, so anyways, have you learned squeeze theorem?

Answer 26

Okay, so, now that that tangent has been gotten out of our systems :)

Answer 27

Maybe I have, but I've never heard of it by that name? I'm not sure.

Answer 28

it can also be called sandwich or something like that

Answer 29

if not we can find another way I'm sure

Answer 30

Can you maybe just show me the first step? Because I did the conjugate and got x-1. I came out with the correct answer, but the first step is marked wrong.

Answer 31

ok, so what conjugate did you use?

Answer 32

show me that step

Answer 33

I used \[\sqrt{x-1}\]

Answer 34

ok can you show me how you did it? It might just be a quick fix then

Answer 35

I think we can do this with algebra

Answer 36

One thing I like to do is re-parametrize... so I let \(u=x-3\) and \(x=u+3\), because \(0\) limits seem easier to work with.

Answer 37

that is true, but wait a sec, he is showing his steps. Although I do think squeeze is the easiest once learned

Answer 38

\[\lim_{x \rightarrow 3} \frac{ 1 }{ x-3 }(\frac{ 1 }{ \sqrt{x+1} }-\frac{ 1 }{ 2 })\]
Becomes: \[
\lim_{u\to 0}\frac{1}{u}\left(\frac{ 1 }{ \sqrt{u+4} }-\frac{ 1 }{ 2 }\right)
\]

Answer 39

\[(\frac{ 1 }{ \sqrt{x+1} }-\frac{ 1 }{ 2 }) \times \frac{ \sqrt{x-1} }{ \sqrt{x-1} } = \frac{ \sqrt{x-1} }{ x-1 } - \frac{ \sqrt{x-1} }{ 2 \times \sqrt{x-1} }\]

Answer 40

No, conjugates don't work that way.

Answer 41

That's what my teacher said. My question is why.

Answer 42

The conjugate of \(\sqrt{x+1}\) is just \(\sqrt{x+1}\).
However, the conjugate of \(\sqrt{a}+b\) is \(\sqrt{a}-b\).

Answer 43

Ah. Okay.

Answer 44

yea, what he said, got there before I could

Answer 45

@FibonacciChick666 what is the name for what you are referring to as the "squeeze theorem"?

Answer 46

squeeze theorem is it's name. It is the most common name used

Answer 47

You guys are just heading in all types of directions.

Answer 48

Maybe reign it in a little bit.

Answer 49

no, but wio is right about that step.

Answer 50

Yes, I see that. I know this. However, what is the right way to go about it?

Answer 51

although I was extremely happy to see you did multiply by a fraction equivalent to 1

Answer 52

I think it will be less confusing if you start with this:\[
\lim_{u\to 0}\frac{1}{u}\left(\frac{ 1 }{ \sqrt{u+4} }-\frac{ 1 }{ 2 }\right)
\]

Answer 53

....

Answer 54

^that too.

Answer 55

And the thing I would want to do first is get rid of fractions by multiplying by the lcm. In this case it would be \(2\sqrt{u+4}\).

Answer 56

I agree,

Answer 57

Okay. Never mind. Someone else showed me the right way to do it.

Answer 58

there are like 10 ways to go about this

Answer 59

not just one correct way

Answer 60

"the" right way, lol

Answer 61

And I just needed 1.

Answer 62

Wait, how many people are helping you with this problem at the same time?

Answer 63

Too many, apparently lol. But thanks for the generosity in trying to help.

Answer 64

well, I tried giving you one, you didn't care for it. Now wio is giving you another one, which is a very nice way, and you are disregarding it.

Answer 65

I'm not disregarding it.

Answer 66

Please, try not to misinterpret my intentions.

Answer 67

@wio do you have any idea what he means by the "squeeze theorem?"

Answer 68

Wio's method he is about to give is probably the most elegant algebraic way

Answer 69

I don't need elegant. I just needed a direction to go in, and you guys took the most roundabout way of getting there.

Answer 70

@triciaal https://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/squeeze-theorem
http://en.wikipedia.org/wiki/Squeeze_theorem
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/squeezedirectory/SqueezePrinciple.html

Answer 71

Squeeze theorem is \[
g(x)\leq f(x)\leq h(x)
\] where \[
\lim g(x)=L,\quad \lim h(x) = L
\]

Answer 72

I'm not disregarding anything. I just wanted an explanation as to why my way was incorrect. You guys gave it to me. However, more questions were added into the mix than necessary.

Answer 73

not intentionally lol, but after all, math happens :)

Answer 74

The people on this site used to be a lot nicer and a lot more focused.

Answer 75

and anyways, I think you should look at wio's method. It is much simpler than the way you were going about it.

Answer 76

let \(f(x)=\dfrac{1}{\sqrt{x+1}}\)

then \[\lim_{x \to3} \frac{ 1 }{ x-3 }(\frac{ 1 }{ \sqrt{x+1} }-\frac{ 1 }{ 2 })=\lim_{x\to3}\frac{f(x)-f(3)}{x-3}=f'(3)\]

Answer 77

I suppose that does work when you have \(0/0\) forms, hmmm

Answer 78

@Zarkon love that, i din't even see that relation too

Answer 79

Thank you, Zarkon. This is the way that I'm used to. Thank you. This is how people should answer questions. Just quick and to the point.

Answer 80

no, that is just giving an answer and not what this site is about. He only gave it to show yet another way of solving this problem

Answer 81

@jayy2014 We don't even know that you know the derivative though...

Answer 82

if you don't know squeeze theorem, you don't know derivative

Answer 83

I can find it. I'm not new to Calculus. I just needed a refresher.

Answer 84

This site isn't about giving answers; I know this. I'm not new here.

Answer 85

which is why, you had 3 views and 3 ways to go about this problem

Answer 86

But you guys were taking too long to tell me about those three ways.

Answer 87

You were responding to everyone at the same time. It's like a teacher trying to answer three questions at once. Someone is going to leave unsatisfied.

Answer 88

... this is math not McDonald's. It takes time to learn and understand the material. And with group answering that happens. then listen to wio, he has a more streamlined style

Answer 89

Either way, I see where I made my mistake, and it was a pretty simple one. I'm surprised at myself. Thanks for the help, as unorganized as it was. I'm done with this problem, and this conversation.

Answer 90

Math, not McDonald's. Pfft. I'm done.

Answer 91

@wio and @Zarkon thank you
@jayy2014 sorry if I invaded your space

Answer 92

Though Zarkon's Method is elegant,
the first though which comes to mind while seeing something in sqrt is to rationalize :)