Question

To evaluate this limit:

lim sqroot(4-x)-2
x-0 ------
x

Could you not simply re-arrange the numerator to get rid of everything except the x, since sqroot4-2 would equal 0, leaving the -x over x?

Answer 1

The example in the book says I must conjugate, but this seems simpler if there is a way to get around it ending up in -x/x.

Answer 2

\[\large \lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x}\]

like this ?

Answer 3

Yes, exactly.

Answer 4

I'm still wondering how do you think the numerator can be simlified to \(\large -x\) without doing the conjugate thingy hmm

Answer 5

Notice entire \(\large 4-x\) is inside squareroot
not just \(\large 4\)

Answer 6

Must they be taken as one number then? It cannot be both √4 and √-x?

Answer 7

you cannot break squareroot like that

Answer 8

Ah, my mistake then

Answer 9

\[\sqrt{a-b} \ne \sqrt{a}-\sqrt{b}\]

Answer 10

I assume I must conjugate to evaluate this then?

Answer 11

thats right but how does a conjugate help here and why is it required here ?

Answer 12

why can't we simply plugin x = 0 ?

Answer 13

Would it not then be indeterminate?

Answer 14

Exactly! so we try to change the form by multiplying it by a special kind of 1 :
\[1 = \dfrac{\sqrt{4-x}+2}{\sqrt{4-x}+2}\]

Answer 15

that magically cancels the bad stuff and allows you to plugin x = 0 and get a determinate form

Answer 16

The book I am using teaches this method as conjugation, though I've never seen this term used in this manner before.

Answer 17

\[\large \begin{align}
\lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x} &= \lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x}\times \dfrac{\sqrt{4-x}+2}{\sqrt{4-x}+2} \\~\\
&= \cdots
\end{align}\]

Answer 18

yes "conjugate" is a name which has similar/different meanings in different contexts

here conjugate just refers to changing the sign of middle term in a binomial

Answer 19

for example

conjugate of \(2-\sqrt{3}\) is \(2+\sqrt{3}\)

Answer 20

I am currently stuck on reconciling the denominator
\[x(\sqrt{4-x}+2)\] to be the same as the numerator
\[\sqrt{4-x}^2-2^2\]

Answer 21

At least, I assume that is the next step.

Answer 22

you're doing good with the numerator, keep going..

Answer 23

The good old "Let's multiply by a fancy form of 1" trick to change your perspective to make the problem to be easier strikes yet again. =D

Answer 24

Now I have\[\frac{ 4-x-4 }{ }\], but still no idea how to get the denominator to match

Answer 25

And I'm not too sure of the end result I've come up with on the numerator as well

Answer 26

:) have faith and keep going..

Answer 27

I guess it would be 4-x+4. since -2^2 is +4...

Answer 28

no,it's not (-2)^2 it's -2^2

Answer 29

4-x-4 simplifies to -x yes ?
next look at denominator, there is a x waiting to get cancelled with the numerator x

Answer 30

But the \[\sqrt{4-x}\] cannot be simplified with just the x, can it?

Answer 31

\[\lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x} = \lim_{x \rightarrow 0} \frac{ -1 }{ \sqrt{4-x} + 2 }\]

Answer 32

now what's ur idea? do u think you can just put x=0 in that and find it?

Answer 33

I apologize but I cannot quite see where I'm going wrong here.

Answer 34

@ganeshie8 , sry u were helping out ;(

Answer 35

@Jaeriko , what have u got so far?

Answer 36

At the moment, my equation looks like this:

\[\frac{ \sqrt{4-x}^2 -2^2}{ x(\sqrt{4-x } +2)}\]

Answer 37

correct ;)

Answer 38

Unsure of whether the numerator will simplify to -x or x, and completely in the dark as to how to simplify the denom.

Answer 39

I do not understand the difference between -2^2 and (-2)^2, for the numerator issue.

Answer 40

As both are -2 by -2, as I understand it.

Answer 41

well, \(-a ^{2} = (-1)(a^{2})\)

Answer 42

thats a good question

Answer 43

u see in that equation we have \( -2^2 \)

Answer 44

it means you should first find \( 2^2 \) then make it negative.

Answer 45

I understand what you're saying there, I've just never heard of it used like that before.

Answer 46

Nor have I ever run into an equation where it was not simply (-x)x(-x) in a -x^2 situation. Is there a name for this rule I can look up?

Answer 47

no i havn't heard name of a rule for it...you know i DIDN'T catch what ur saying.please state WHAT you have problem with...you don't know how you can find the numerator?

Answer 48

let me ask you a question @Jaeriko

whats the value of \(\large 3^2 - 2^2\) ?

Answer 49

I do not understand how I can simplify the current state of my equations numerator to the -1 that you and the book state is the solution.

Answer 50

5, if I understand correctly.

Answer 51

Right! how did you get 5 ?

Answer 52

Ah, I see

Answer 53

I suppose I've done this many times without truly noticing, then

Answer 54

\[3^2 - 2^2 = 9 - 4 = 5\]

similarly

\[(\sqrt{4-x})^2 - 2^2 = 4-x - 4 = -x\]

Answer 55

@ganeshie8 nice explanation ;)

Answer 56

its bit hard to explain.. but i see you figured it out already :)

Answer 57

Okay, I understand that step now...but what about turning -x into -1?

Answer 58

Or does that come later?

Answer 59

that comes later
-x wont change into 1 just like that without some balck magic haha!

Answer 60

well \(\frac{ -x }{ x } = -1\)

Answer 61

Okay, so how do we get eliminate the \[x( \sqrt{4-x}+2)\] on the bottom now? I don't know how we can do anything with the \[\sqrt{4-x}\], as we have to deal with them as a group instead of individual numbers.

Answer 62

Can we have \[\sqrt{4-x^2}\] now?

Answer 63

its much simpler than that
the only bad guy in the denominator is \(x\) because it pulls the denominator to 0

Answer 64

just cancel the x in denominator with the numerator x
then plugin x = 0

you're done!

Answer 65

here everything at one place :
\[\require{cancel}\large \begin{align}
\lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x} &= \lim\limits_{x\to 0} \, \dfrac{\sqrt{4-x}-2}{x}\times \dfrac{\sqrt{4-x}+2}{\sqrt{4-x}+2} \\~\\
&= \lim\limits_{x\to 0} \,\dfrac{(\sqrt{4-x})^2 - 2^2}{x(\sqrt{4-x}+2)} \\~\\
&= \lim\limits_{x\to 0} \,\dfrac{4-x-4}{x(\sqrt{4-x}+2)} \\~\\
&= \lim\limits_{x\to 0} \,\dfrac{-x}{x(\sqrt{4-x}+2)} \\~\\
&= \lim\limits_{x\to 0} \,\dfrac{-\cancel{x}}{\cancel{x}(\sqrt{4-x}+2)} \\~\\
&= \lim\limits_{x\to 0} \,\dfrac{-1}{\sqrt{4-x}+2} \\~\\
&=\dfrac{-1}{\sqrt{4-0}+2} \\~\\
&=\dfrac{-1}{2+2} \\~\\
&=\dfrac{-1}{4}
\end{align}\]

Answer 66

Oh! That's amazing, I didn't even see it coming for some reason. Thank both of you so much for your help!

Answer 67

As soon as I realized you could cancel the -x and x with each other to leave the -1, it feel into place.

Answer 68

I wish I had but more medals to give

Answer 69

yes you get use to these after working one two examples and will feel its really easy
yw :)

Answer 70

ur welcome @ganeshie8 helped you well for his awesome explanation....

@Jaeriko wait a moment,i have prepared some awesome problems,you can do them for fun and check your understanding

Answer 71

\(\large \lim_{h \rightarrow 0} \frac{ \sqrt{1+h} - 1}{ h }\)

\(\large \lim_{x \rightarrow -4} \frac{ \sqrt{x^2 + 9} -5 }{ x+4 }\)

\(\large \lim_{x \rightarrow 16} \frac{ 4-\sqrt{x} }{ 16x - x^2}\)

\(\large \lim_{x \rightarrow 7} \frac{ \sqrt{x+2}-3 }{ x-7 }\)

Answer 72

I greatly appreciate the time you have both spent explaining this to me.

Have a good night, and thank you again.

@PFEH.1999 ,I'm writing those down for practice after I finish work tomorrow. Thank you so much for your time and effort!

Answer 73

Nice :) All of these can be worked out using conjugate thingy..
also they all can be evaluated by using definition of derivative !