Question

\[\lim_{x \rightarrow \infty} ((x+1)(x+2)(x+3)(x+4))^{1/4}-x\]

Answer 1

@ganeshie8 ?

Answer 2

@hartnn ?

Answer 3

@DLS ?
@Azureilai ?
@bcr1997 ?
@chmvijay ?
@Hero ?
@insa ?
@karlacampos ?
@matricked ?

Answer 4

should i tell my approach?

Answer 5

Go ahead.

Answer 6

\[\lim_{x \rightarrow \infty}x[(1+1/x)+(1+2/x)+(1+3/x)(1+4/x)]^{1/4}-x\]

Answer 7

sorry i think i wrote wrong instead of + sign there should be multiplication

Answer 8

\lim_{x \rightarrow \infty}x[(1+1/x)(1+2/x)(1+3/x)(1+4/x)]^{1/4}-x

Answer 9

Perfect!

Now,

\[\lim_{x \rightarrow \infty} {(((x+1)(x+2)(x+3)(x+4))}^{\frac{1}{4}} - x)\]

Now after you take x common from all the terms, you have \((x^4)^{\frac{1}{4}} = x\)

Then you'd get:

\[\lim_{x \rightarrow \infty} x((1 + \frac{1}{x})(1 + \frac{2}{x})(1 + \frac{3}{x})(1 + \frac{4}{x}) - 1)\]

Use the property of limits: \[\lim_{x \rightarrow a} (f(x) ~~.~ ~g(x)) = \lim_{x \rightarrow a} f(x) ~~. ~~\lim_{x \rightarrow a} g(x)\]

You'll get: \[\lim_{x \rightarrow \infty} x ~~. ~~\lim_{x \rightarrow \infty} ((1+ \frac{1}{x})(1 + \frac{2}{x})(1 + \frac{3}{x})(1 + \frac{4}{x}) - 1)\]

Which gives you: \(\infty . 0\) = 0 ?

Something does not seem right or is it?

@hartnn
@ganeshie8

Answer 10

I missed out the \(\frac{1}{4}\) above.
Even if it was there, it wouldn't have made any difference I suppose.

Answer 11

I am very sure, 0 is not the answer. Hold on let me try again.

Answer 12

if it helps, wolf says 5/2
http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Crightarrow+%5Cinfty%7D++%28%28%28x%2B1%29%28x%2B2%29%28x%2B3%29%28x%2B4%29%29%5E%7B1%2F4%7D-x%29

Answer 13

Do we have to use binomial Theorem by any chance?
Yeah, even I checked on wolfram, but couldn't check the step-by-step solution.

Answer 14

got a headach trying to expand the LH @_@

Answer 15

Got the answer? O.O

Answer 16

\[\lim_{x \to \infty} ~\left((x+1)(x+2)(x+3)(x+4)\right)^{1/4}-x\]

\[\lim_{t \to 0} ~ \dfrac{\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{1/4}-1}{t}\]

indeterminate form, apply L'hosps

\[\lim_{t \to 0} ~ \dfrac{1}{4}\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{-3/4}\times \left(1 + 10t + \mathcal{O(t^2)} \right)'\]

\[\lim_{t \to 0} ~ \dfrac{1}{4}\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{-3/4}\times \left(0 + 10 + \mathcal{O(t)} \right)\]

\[ \dfrac{1}{4}\left((1+0)(1+0)(1+0)(1+0)\right)^{-3/4}\times \left(0 + 10 + 0 \right)\]

\[\dfrac{5}{2}\]

Answer 17

Awesome! :'O

Answer 18

time to use this
http://assets.openstudy.com/updates/attachments/53b23d16e4b09f140ca3dc35-geerky42-1404190111618-tumblr_n75mv1a8ko1son60zo1_400.gif

Answer 19

hahahahahahah! :D

Answer 20

me :- one step late *duh*

Answer 21

lol that suggestion of t->0 was partly given by @DLS

Answer 22

* going to kill myself *

lolz , it really cuz a headach xD

Answer 23

need to review calcules ,no doubt
thats it xD

Answer 24

I wanted to use, L'Hospital. :'|

Answer 25

<3 <3

Answer 26

@ganeshie8

Answer 27

how did u transform it to the 2nd line

Answer 28

@ganeshie8 ?

Answer 29

substitute 1/x = t

Answer 30

\[\lim_{x \rightarrow \infty} ((1+\frac{ 1 }{ t })(1+2/t)(1+3/t)(1+4/t))^{1/4}-1/t\]

Answer 31

added few more lines :


\[\lim_{x \to \infty} ~\left((x+1)(x+2)(x+3)(x+4)\right)^{1/4}-x\]

\[\lim_{x \to \infty} ~x \left[\left((1+\frac{1}{x})(1+\frac{2}{x})(1+\frac{3}{x})(1+\frac{4}{x})\right)^{1/4}-1\right]\]

sub \(\frac{1}{x} = t\)
as \(x \to \infty , t \to 0\)

\[\lim_{t \to 0} ~ \dfrac{\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{1/4}-1}{t}\]

indeterminate form, apply L'hosps

\[\lim_{t \to 0} ~ \dfrac{1}{4}\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{-3/4}\times \left(1 + 10t + \mathcal{O(t^2)} \right)'\]

\[\lim_{t \to 0} ~ \dfrac{1}{4}\left((1+t)(1+2t)(1+3t)(1+4t)\right)^{-3/4}\times \left(0 + 10 + \mathcal{O(t)} \right)\]

\[ \dfrac{1}{4}\left((1+0)(1+0)(1+0)(1+0)\right)^{-3/4}\times \left(0 + 10 + 0 \right)\]

\[\dfrac{5}{2}\]

Answer 32

what is this (1+10t ....(t^2) ??

Answer 33

expand the product \((1+t)(1+2t)(1+3t)(1+4t)\)

Answer 34

Clearly, the constant term is 1x1x1x1 = 1

Answer 35

the coefficient of t = (1 + 2 + 3 + 4) = 10

Answer 36

we don't need higher order terms cuz they vanish in the next line when u take the limit.

Answer 37

if thats confusing, you may simply expand the whole thing and take the derivative ^

Answer 38

thanks i got it!

Answer 39

:)