Question

What are the steps for:

lim x->∞ (√(x^2+4x+1)-x)

Thanks

Answer 1

Is this the question?
\[\color{red}{\lim_{x \rightarrow \infty}(\sqrt{(x^2+4x+1)}-x.)}\]

Answer 2

yes

Answer 3

Ive rationalized it but I am stuck on where to go from there

Answer 4

\[4x+1/\sqrt{x^2+4x+1}+x\]

Answer 5

So carefully note that it is a type of question of limit At \[\pm \infty.\]with a radical in it.
So If i do two examples for u .
Then, would u be able to do ur question?

Answer 6

?

Answer 7

solve away

Answer 8

I was just confused about the plus/minus you had but I realized it was because of the radicand

Answer 9

so what do u want now
examples or ur question's solution.
former one is better.

Answer 10

examples would be great. thanks

Answer 11

k! suppose if we take this as first example: -
\[\color{red}{\lim_{x \rightarrow \infty}\frac{x+3}{\sqrt{x^2+4}}.}\]k:)

Answer 12

alright

Answer 13

Now, If we notice the highest power of 'x' in the denominator; we see it is \[\sqrt{x^2}=x^1\]
clear upto this?

Answer 14

yea

Answer 15

now the next step would be as: -
what to do of highest power of x is this.
\[\lim_{x \rightarrow \infty}{\frac{\frac{1}{x}(x+3)}{\frac{1}{x}\sqrt{x^2+4}}}\]
clear?

Answer 16

yes

Answer 17

k! then
u can write the above as
\[\LARGE{\lim_{x \rightarrow \infty}{\frac{1+\frac{3}{x}}{\sqrt{\frac{1}{x^2}}\sqrt{x^2+4}}}}\]clear?

Answer 18

yep

Answer 19

now it is nothing but this: -
\[\lim_{x \rightarrow \infty}{\frac{1+\frac{3}{x}}{\sqrt{1+\frac{4}{x^2}}}}\]

Answer 20

clear?

Answer 21

yes

Answer 22

then if u apply limit here i mean that
If u put x->infinite here
then u will see this\[\color{blue}{\frac{3}{x}=0}\] and\[\color{blue}{\frac{4}{x^2}=0.}\]

Answer 23

clear?

Answer 24

yes

Answer 25

so it becomes 1/1=1

Answer 26

then what u left with is this\[\large{\color{purple}{\frac{1}{\sqrt{1}}=1}}\]

Answer 27

ya u gt right.

Answer 28

but what happens when you have no denominator like my example?

Answer 29

then u just make denominator by conjugate method.
clear?

Answer 30

any doubt?

Answer 31

yea Im afraid so. I understood the example you showed me. I had been taught that method already. But with the problem I posted earlier, I can't seem to apply it even after I used the conjugate and rationalize the function.

Answer 32

wait for upto 5 minutes.
first I will try that.
k!

Answer 33

yes

Answer 34

is ur answer \[\color{green}{\infty?}\]

Answer 35

no the correct answer is 2

Answer 36

what!

Answer 37

http://www.wolframalpha.com/input/?i=lim+x-%3E ∞+%28√%28x%5E2%2B4x%2B1%29-x%29

Answer 38

i don't think that link works, but if you plug it into wolfram alpha, it will show you the answer and the steps, but I was trying to find a more concise way of doing the problem. Usually wolfram uses a cumbersome approach

Answer 39

ya:)
If I do what I did in my copy
would u be able to find my error?

Answer 40

only i say try!

Answer 41

doubt it. I can't figure it out myself.

Answer 42

k! np:)
u see just what i did?

Answer 43

no i didn't

Answer 44

I gt after rationalising\[\lim_{x \rightarrow \infty}{\frac{\frac{1}{x}(4x+1)}{{\sqrt{\frac{1}{x^2}}(\sqrt{x^2+4x+1}+x).}}}\]

Answer 45

ok

Answer 46

ahhh I see it.

Answer 47

I didn't see how you could cancel out 1/x and 1/x^2. I didn't take into account that the radicand reduced it to 1/x

Answer 48

& then just multiplied it whole
\[\lim_{x \rightarrow \infty}{\frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+x}}\]& applied the limit & gt
\[\frac{4+0}{\sqrt{1+0+0}+\infty}.\]\[\LARGE{\color{red}{\frac{4}{\infty}=\infty.}}\]

Answer 49

&