Question

Calculus Question:
Topic : Limits

please wait for me to write my question. Thank you

Answer 1

\[\lim_{x \rightarrow \infty} (x^2 +4)^{1/2} - (x^2-1)^{1/2}\]

In the above it is the question. We have to rationalize since inifinity cannot be subtracted

Answer 2

\[\sqrt{x^2+4}-\sqrt{x^2-1}\times \frac{ \sqrt{x^2+4} +\sqrt{x^2-1}}{ \sqrt{x^2+4}+\sqrt{x^2-1} }\]

Answer 3

The answer turned to be 5/ (x^2+4)^1/2 + (x^2-1)^1/2 = 0

Answer 4

put abract also in the first portion.

Answer 5

why did u square it ?

Answer 6

nevermind, I got it to the power of 1/2 is square root of something. you are right

Answer 7

\[\sqrt{x}=x ^{\frac{ 1 }{ 2 }}\]

Answer 8

yes

Answer 9

the problem is I don't know what to do after I raionalize

Answer 10

write it\[\left( \sqrt{x^2+4}-\sqrt{x^2-1} \right)\]

Answer 11

put \[x=\frac{ 1 }{ y }\]
\[y \rightarrow0,~as~x \rightarrow \rightarrow \infty \]

Answer 12

ok, but my book gave this answer.

\[\frac{ 5 }{ (x^2+4)^{1/2} +(x^2-1)^{1/2} }\]

Answer 13

I don't know how to get this.

Answer 14

@sshayer what should we do next?

Answer 15

if you want to rationalize only then that is the correct answer.
if you want to find limit then proceed as i have suggested above.

Answer 16

i want to find the limit

Answer 17

put x=1/y as suggested above.

Answer 18

\[((1/y))^2 +4 )^1/2 - ((1/y)^2-1)^1/2\]

Answer 19

ok what should i do next?

Answer 20

\[\frac{ 5y }{\sqrt{1+4y^2}+\sqrt{1-y^2}}\]

Answer 21

after rationalizing put x=1/y

Answer 22

no

Answer 23

\[now~ limit ~is~y \rightarrow 0\]

Answer 24

ok can u do me a favor. Show me what you mean step by step. then I will ask u for clarification of any step I don't understand. Now I am confused.

Answer 25

actually you should get after you substituting
i have solved above.

Answer 26

I am sorry my friend. This is not the way someone explains something to someone else who does not understand a new topic. You say substitute this and then substitute the next thing for something else.

If I were you,
I would write all the steps since I have the answer already, and then the person who is seeking for help would ask you to explain one step that he does not understand.

I am just giving a suggestion. I hope you understand what I mean.

Answer 27

when you put y=0
\[\lim_{y \rightarrow 0}\frac{ 5y }{ \sqrt{1+4y^2} +\sqrt{1-y^2}}=\frac{ 5*0 }{ \sqrt{1+0}+\sqrt{1-0} }=\frac{ 0 }{ 2 }=0\]

Answer 28

I don't understand how u got 5y / square root 1 + 4y^2 + square root 1-y^2

Answer 29

1. your first step is rationalize.
2.second step is put x=1/y
3. find the limit as y approaches 0

Answer 30

\[\frac{ 5 }{ \sqrt{\frac{ 1 }{ y^2 }+4}+\sqrt{\frac{ 1 }{ y^2 }-1} }=\frac{ 5 }{ \sqrt{\frac{ 1+4y^2 }{ y^2 }}+\sqrt{\frac{ 1-y^2 }{ y^2 }} }\]
\[=\frac{ 5 }{ \frac{ \sqrt{1+4y^2} }{ y }+\frac{ \sqrt{1-y^2} }{ y } }=\frac{ 5 }{ \frac{ \sqrt{1+4y^2}+\sqrt{1-y^2 } }{ y } }\]
\[=\frac{ 5y }{ \sqrt{1+4y^2}+\sqrt{1-y^2} }\]

Answer 31

perfect! :)

Answer 32

just my last question is:

why did my book give the answer:

\[\lim_{x \rightarrow \infty} \frac{ 5 }{ (x^2+4)^1/2 + (x^2-1)^1/2 } = 0\]

why is there all those variables at the bottom?qq

Answer 33

I mean is there any rule or law I can know to know the answer is immediately 0.

Answer 34

it is not always zero.
But when the limit is infinity try to bring the variable in the denominator.

Answer 35

dayumm im studying the same thing rn...ded

Answer 36

@sshayer thank you for your help. you spent a long time :)