# How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?

Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$

I have tried rationalizing but there is no pattern that I can observe.

Edit:

So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$

$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$

• Just a thought - how about making the change of variables $w = \frac{1}{x}$ and using the binomial series on the resulting limit of $w$ tending to 0 for the expression $\frac{1}{w^2} ((1+w^2)^{\frac{1}{2}} - (1+w^3)^{\frac{1}{3}})?$ [I am not suggesting that this is an answer; I just want to know if my thinking here is correct.] – SystematicDisintegration Dec 12 '17 at 0:57
• Try the appropriate conjugate for $6$th powers. Hint: factor $x^6-y^6$. – Michael Burr Dec 12 '17 at 1:09

You can do it with Taylor expansions:

$x\left(1+x^{2}\right)^{1/2}-x\left(1+x^{3}\right)^{1/3}= x^{2}\left(1+\frac{1}{x^{2}}\right)^{1/2}-x^{2}\left(1+\frac{1}{x^{3}}\right)^{1/3} =x^{2}\left(1+\frac{1}{2}\cdot\frac{1}{x^{2}}+o\left(\frac{1}{x^{2}}\right)\right)-x^{2}\left(1+\frac{1}{3}\cdot\frac{1}{x^{3}}+o\left(\frac{1}{x^{3}}\right)\right) =\frac{1}{2}-\frac{1}{3x}+o\left(1\right) \to\frac{1}{2}$

• I guess the approach I posted as a comment was right then...just a couple of seconds late though! – SystematicDisintegration Dec 12 '17 at 0:59

I like the Taylor expansion approach, but wondered if there was a more "Cal I" approach.

Using $$A-B = (A^{1/6} - B^{1/6})(A^{5/6} + A^{4/6}B^{1/6} + A^{3/6}B^{2/6} + \cdots + B^{5/6})$$ with $A = (x^2+1)^3$ and $B = (1+x^3)^2$ gives $$\begin{multline*} (x^2+1)^{1/2} - (x^3+1)^{1/3}\\ =\frac{(x^2+1)^3 - (x^3+1)^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}}\\ =\frac{3x^4+2x^3+3x^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}} \end{multline*}$$ Now $$\begin{multline*} x((x^2+1)^{1/2} - (x^3+1)^{1/3})\\ =\frac{x^5(3+2/x+3/x^2)}{x^5((1+1/x^2)^{5/2}+ (1+1/x^2)^{2}(1+1/x^3)^{1/3} + \cdots + (1+1/x^3)^{5/3})}\\ \end{multline*}$$

which goes to $3/6$ as $x\to \infty$.

• And $3/6$ approaches $1/2$ quite quickly.... – CiaPan Dec 12 '17 at 8:14
• @CiaPan :-) ... – Malcolm Dec 12 '17 at 8:25

Let's put $x=1/t$ so that $t\to 0^{+}$ and the expression under limit is transformed into $$\frac{\sqrt{1+1/t^{2}}-\sqrt[3]{1+1/t^{3}}}{t}=\frac{\sqrt{1+t^{2}}-\sqrt[3]{1+t^{3}}}{t^{2}}$$ and this can be expressed as a difference of two standard limits $$\lim_{t\to 0^{+}}\frac{(1+t^{2})^{1/2}-1}{t^{2}}-t\cdot \frac{(1+t^{3})^{1/3}-1}{t^{3}}$$ which equals $1/2-0(1/3)=1/2$. The standard limit formula $$\lim_{x\to 0}\frac{(1+x)^{n}-1}{x}=n$$ used above is a direct consequence of the more general limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$

$$x^2[(1+1/x^2)^{1/2} - (1+1/x^3)^{1/3}].$$
Now use the fact that $(1+h)^p = 1 + ph +o(h)$ as $h\to 0.$ (This fact is equivalent to the statement that the derivative of $(1+x)^p$ at $x=1$ is $p.$)
This limit would be equal to $1/2$. See my demo :
$$\lim_{x \to \infty} {x\sqrt{x^2+1}-x (x^3+1)^{1/3}} = \lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})}$$ Now, let us see how much is each member, by Laurent-Taylor expansion : $$\sqrt{x^2+1} = x+ {1 \over {2x}} - {1 \over {8x^3}}+{1 \over {16x^5}}-...$$ $$(x^3+1)^{1/3} = x+ {1 \over {3x^2}} - {1 \over {9x^5}}+{5 \over {81x^8}}-...$$ Resting this 2 members : $$\sqrt{x^2+1}-(x^3+1)^{1/3} = {1 \over {2x}}-{1 \over {3x^2}} - {1 \over {8x^3}}+({1 \over 16}+{1 \over 9}){1 \over x^5}-...$$ Now we substitute this in our limit : $$\lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})} = \lim_{x \to \infty} { x ({1 \over {2x}}-{1 \over {3x^2}}-...)}= {1 \over 2}$$