Question

can you help me with that :
lim x2−−√3(x+1−−−−−√−x2−1−−−−−√) as x approch to infinity

Answer 1

I'm confused as to what is happening in that equation.

Answer 2

can you help me with that :
lim\[\sqrt[3]{x^{2}}(\sqrt{x+1}-\sqrt{x^{2}-1})\] as x approaches to infinity @Bmannn13

Answer 3

Sure!

Answer 4

let's look at the limit (sqrt{x+1}-sqrt{x^2-1}) since that one decides the whole thing

Answer 5

so what do you think we can do here?

Answer 6

any possible technique that we can use to get rid of indeterminate form

Answer 7

I just got the answer, but someone else is helping already!

Answer 8

oh no answers :)
just help

Answer 9

let the person do the work :)

Answer 10

just write it please ! @Bmannn13

Answer 11

@M_lowreen it is not good for you ^_-
you gotta do the work my dear

Answer 12

we can walk you through if you have the interest

Answer 13

it's 00:20 am and i've studying for more than 6 hours.... can anyone help me please !

Answer 14

That's true! I wasn't just going to give you the answer either.

Answer 15

i know... i need just an idea !

Answer 16

does not matter! you still gotta do your own assignment by your own!

Answer 17

Here, do this:

Answer 18

\[d/dx(\sqrt{x+1}-\sqrt{x ^{2}-1})\]

Answer 19

what does d/dx mean ? @Bmannn13

Answer 20

Find the derivative.

Answer 21

i haven't learned this yet! :/

Answer 22

okay so when we are faced with this sort of expression with radicals
we can think of multiplying by "forget the word lol"
just gonna give you an example \[(\sqrt{a}-\sqrt{b}) \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}\]

Answer 23

you can do that algebraic manipulation to get rid of indeterminate problem

Answer 24

surely you LEARNED THAT ONE ^_-

Answer 25

DARN MY MIND CAN'T RECALL THE NAME LOL

Answer 26

ok, thanks ! I'll try it now :) :) @xapproachesinfinity

Answer 27

we do the same for our problem here

Answer 28

a is this case x+1
b is x^2-1

Answer 29

the game is to have some cancellations to make it more easier to study its limit

Answer 30

I'm exchausted.... i;ve been studing since 5 pm..... tomorrow i'm going to take an advanced level exam @xapproachesinfinity

Answer 31

the word i was looking for it the conjugate
so we multiply and divide by the conjugate
darn the word didn;t want to come out hahaha

Answer 32

so what does that have to do with this! hahaha

Answer 33

oww ... i know this !

Answer 34

I may sound hard on ya, but believe that won't help any good!

Answer 35

okay so we have \[\lim \sqrt{x+1}-\sqrt{x^2-1}=\lim (\sqrt{x+1}-\sqrt{x^2-1})\frac{\sqrt{x+1}+\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^2-1}}\]

Answer 36

first step

Answer 37

then \[=\lim\frac{(\sqrt{x+1}-\sqrt{x^2-1})(\sqrt{x+1}+\sqrt{x^2-1})}{\sqrt{x+1}+\sqrt{x^2-1}}\]

Answer 38

I just rewrite so it can appear to you as \[(a-b)(a+b)\]

Answer 39

but \[\sqrt{x^{2}-1}=\sqrt{(x-1)(x+1)}\]

Answer 40

which you should now since you took algebra classes

Answer 41

nope

Answer 42

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

Answer 43

yeah,, i know right... up to here ... but then?

Answer 44

you should already know that \[(a-b)(a+b)=a^2-b^2\]
i used this
since you are in calc class this has to be stuck in your memory

Answer 45

i know ittt

Answer 46

earlier you said you can't do it now you saying you know???

Answer 47

then why you didn't do it?

Answer 48

i said i know this algebra formulas ,,, but they confuse me all

Answer 49

if they confuse then you don't know them yet lol

Answer 50

just kidding :)

Answer 51

Problem solved *___* lol

Answer 52

so back to your problem
\[=\lim \frac{x+1-x^2+1}{\sqrt{x+1}+\sqrt{x^2-1}}\]

Answer 53

simplify more

Answer 54

what if i factorise at the denominator \[\sqrt{x+1}(1+\sqrt{x-1})\] ? @xapproachesinfinity

Answer 55

how is that possible ? cannot be done
what you did has no meaning
try to distribute and see if you get back to that denominator

Answer 56

i was talking about the top simplify more
i meant just add 1+1
to get 2+x-x^2

Answer 57

at this level it is still indeterminate form oo/oo which is no good

Answer 58

so we have to do something more to bring it to a more simplified form

Answer 59

i know ,, sooo saaadddd

Answer 60

hehe
\[\lim \frac{2+x-x^2}{\sqrt{x+1}+\sqrt{x^2-1}}=\lim\frac{2/x+1-x}{(\sqrt{1/x+1/x^2}+\sqrt{1-1/x^2})}\]

Answer 61

i factored out x top and bottom

Answer 62

now if you tried the evaluate the limit is should be easy

Answer 63

2/x goes to zero
1 to 1
-x goes to -oo so the top by it self goes to -oo
========
the bottom goes to 1
so we have -oo/1 which is -oo

Answer 64

***Hugs*** :)))))) @xapproachesinfinity

Answer 65

you made my day :D

Answer 66

all that work can be done in one scoop if you know about l'hopitals rule

Answer 67

no problem

Answer 68

you may be right.. but it's the first time we work with limits :/ @xapproachesinfinity