Approximation of a function in limit of small and large x

1. Aug 2, 2012

FrankDrebon

Hi all,

I'm having a bit of trouble getting my head round approximations to a function in the limit of small and large values of the x parameter. The function is:

$y = x\left\{ {\left[ {1 + \left( {{1 \over x}} \right)^2 } \right]^{{1 \over 2}} - 1} \right\}$

The paper I'm reading says y becomes independent of x for small values of x. For large values of x, y becomes proportional to (1/x). Has anyone got any ideas of how this can be shown algebraically? I haven't had to do this for a while!

FD

2. Aug 2, 2012

DonAntonio

Let us look at
$$h(x):=\frac{\left(1+\frac{1}{x^2}\right)^{1/2}-1}{\frac{1}{x}}$$

When $\,x\to 0\,$ , both numerator and denominator above go to $\,\pm \infty$ , so applying L'Hospital we get:

$$\lim_{x\to 0}h(x)=\lim_{x\to 0}\frac{-\frac{1}{x^3}\left(1+\frac{1}{x^2}\right)^{-1/2}}{-\frac{1}{x^2}}=\lim_{x\to 0}\frac{1}{x\left(1+\frac{1}{x^2}\right)^{1/2}}=\lim_{x\to 0}\frac{1}{\left(x^2+1\right)^{1/2}}=1$$

the above being possible if $\,x>0\,\,,\,i.e.\,\,\lim_{x\to 0^+}\,$ . If we take the limit from the left just

multiply by $\,-1\,$ after the 3rd. equality above and proceed then as before. The limit then is $\,-1$ .

This shows the paper you're reading is wrong as the value approached definitely depends on
whether x is "very small",

positive or negative . It's easy to see that has x becomes larger and larger the expression in the square root

becomes closer to 1.

Doing exactly the same as above in the case $\,x\to\infty\,$ , we get the limit is now zero, but the

thing about "proporcionality" perhaps means that $\,|h(x)|\leq\frac{M}{x}\,\,\,,\,\,as\,\,\,x \to \infty\,\,,\,M=\,$ a constant. Check this.

DonAntonio

3. Aug 2, 2012

Staff: Mentor

You can calculate the taylor expansion for h(x) at x=0 and for h(1/x) at 1/x=0 - in both cases, use the limit value where h is not well-defined. If x cannot be negative, this limit exists.

Hand-wavy way for small x: (1+z)^(1/2) = 1+z/2+O(z^2) for small z, therefore
$h(x) = x \left(\frac{1}{x}(1+x^2)^{1/2}-1\right) = x\left(\frac{1}{x}(1+\frac{x^2}{2}+O(x^4))-1\right)=1-x+\frac{x^2}{2}+O(x^4)$
Well, I would not call this "independent of x", the single "-x" does not fit to this.
Another way would be to derive your whole equation. If the limit of the derivative is 0 (it is not, unless I made a mistake), you could call this "independent of x".

And small 1/x:
$h(x)=x \left(1+\frac{1}{x^2}+O(\frac{1}{x^4})-1\right) = \frac{1}{x} + O(\frac{1}{x^3})$
Another way here would be to show that y*x approaches a constant for x->infinity.

@DonAntonio: This would be an upper limit, not a proportionality.

4. Aug 5, 2012

Matt Benesi

Isn't that the same as:

$f(x) = x \times \sqrt {1 + \left( {{1 \over x}} \right)^2 } - x$

which is equal to:

$f(x) = \sqrt {x^2 + x^2 \times\left( {{1 \over x^2}} \right) } - x$

which is equal to:

$f(x) = \sqrt {x^2 + 1 } - x$

Which leads us to the conclusions:
$$\lim_{x\to\infty} \sqrt {x^2 + 1 } - x =0$$
and
$$\lim_{x\to 0} \sqrt {x^2 + 1 } - x = 1$$

5. Aug 5, 2012

pwsnafu

x can be negative so the next line is wrong. Remember $\sqrt{x^2} = |x|$