Question

Find the limit

Answer 1

\[\lim_{x \rightarrow \infty}\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3+1}\]

Answer 2

So I multiplied by the conjugate and got:

\[\lim_{x \rightarrow \infty}\ \frac{ x^2 }{ \sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3+1} }\]

Answer 3

How do I go from there?

Answer 4

@ganeshie8 @nincompoop

Answer 5

@jim_thompson5910

Answer 6

careful, its a cube root

Answer 7

Yes I know.

Answer 8

so conjugate thing wont help and you can't really rationalize numerator this way

Answer 9

I can't believe this. I'm taking complex anlysis and I can't do a simple elementary limit >>> .

Answer 10

>.> *

Answer 11

Ohh. What do you suggest then?

Answer 12

because \(\large (\sqrt[3]{a} +\sqrt[3]{b} ) (\sqrt[3]{a} -\sqrt[3]{b} ) \ne a-b \)

Answer 13

Ohh alright.

Answer 14

do u remember (a-b)^3 formula ?

Answer 15

let me google

Answer 16

Yeah I know.

Answer 17

Well sorta.

Answer 18

a^3-b^3=(a-b)(a^2+ab+b^2)

Answer 19

yeah if you multiply the given expression by (a^2+ab+b^2), you can get rid off the cube roots, yes ?

Answer 20

Yes I think I see it.

Answer 21

\[\large \lim_{x \rightarrow \infty}\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3+1} \times \dfrac{(x^3+x^2+1)^{2/3}+(x^3+x^2+1)^{1/3}(x^3+1)^{1/3} + (x^3+1)^{2/3} }{(x^3+x^2+1)^{2/3}+(x^3+x^2+1)^{1/3}(x^3+1)^{1/3} + (x^3+1)^{2/3}}\]

Answer 22

i hate when it happens, it wont fit

Answer 23

basically numerator simplifies to a^3-b^3 so that the radical disappears and you can cancel some unwanted stuff

Answer 24

I can see it if I copy the latex format to wolfram :P .

Answer 25

good idea, but still its annoying >.<

it simplifies to :
\[\large \lim_{x \rightarrow \infty}\dfrac{(x^3+x^2+1) - (x^3+1)}{(x^3+x^2+1)^{2/3}+(x^3+x^2+1)^{1/3}(x^3+1)^{1/3} + (x^3+1)^{2/3}}\]

Answer 26

Okay lets see if I get the same thing...

Answer 27

just apply the formula

Answer 28

Yep, that's what I'm doing haha.

Answer 29

Yep I got your answer :) .

Answer 30

What about the denominator though?

Answer 31

denominator is good

Answer 32

Okay, I should be alright to divde everything by x^3 right?

Answer 33

No wait.

Answer 34

or pull out x^3 from the denominator

Answer 35

Same thing I guess.

Answer 36

It's just bugging me because then the numerator becomes 1/x^(something) which, if I take limit is 0 so the whole thing is 0.

Answer 37

\[\begin{align}\large &\lim_{x \rightarrow \infty}\dfrac{(x^3+x^2+1) - (x^3+1)}{(x^3+x^2+1)^{2/3}+(x^3+x^2+1)^{1/3}(x^3+1)^{1/3} + (x^3+1)^{2/3}} \\~\\&=\lim_{x \rightarrow \infty}\dfrac{x^2}{(x^3+x^2+1)^{2/3}+(x^3+x^2+1)^{1/3}(x^3+1)^{1/3} + (x^3+1)^{2/3}} \\~\\ &=\lim_{x \rightarrow \infty}\dfrac{x^2}{x^2\left[(1+1/x+1/x^3)^{2/3}+(1+1/x+1/x^3)^{1/3}(1+1/x^3)^{1/3} + (1+1/x^3 )^{2/3}\right]} \\~\\ \end{align}\]

Answer 38

x^2 cancels out and you can take the limit

Answer 39

Ohh I see. I was trying to take out an x^3 .

Answer 40

Ahh thank you! I got all that way but I took out x^3 which caused problems.

Answer 41

x^3 becomes x^2 when you take it out of the exponent 2/3

Answer 42

Aha! I see it. :)

Answer 43

And I get 1/3 which is correct I believe. Thanks!

Answer 44

yes wolfram gives the same answer, wonder if there is any other easy way :o

Answer 45

Probably L'hospital's rule?

Answer 46

Wait no. That's a disgusting derivative.

Answer 47

derivatives would be scary with radicals
and moreover u need to put it in indeterminate form which can be tricky

Answer 48

Wolfram won't give steps either >.> .

Answer 49

taylor series may give fast answer

Answer 50

No. I'm writing the solutions to a practice midterm for a first year calculus course and Taylor series comes much later for them.

Answer 51

\[\large \begin{align}

&\lim_{x \rightarrow \infty}\sqrt[3]{x^3+x^2+1}-\sqrt[3]{x^3+1} \\~\\
&\lim_{x \rightarrow \infty}x(1+1/x+1/x^3)^{1/3}-x(1+1/x^3)^{1/3} \\~\\
&\lim_{x \rightarrow \infty}x\left(1+\frac{1}{3}(1/x+1/x^3) + \mathcal{O(1/x^2)}\right) -x\left(1+\frac{1}{3}1/x^3 + \mathcal{O(1/x^2)}\right) \\~\\

&\dfrac{1}{3}
\end{align}\]

Answer 52

its okay, over years you might forget all limit formulas but taylor series always works !