yet another limit...

**Jones** · December 26th 2009, 10:10 AM

Hi,

I have this limit i would like some help with

$\displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{sin(2x)}$

So i tried replacing the sin(2x) term by it's equivalent taylor polynomials for (sin x and cos x)

like so:
$\displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{2\bigg(\frac{x^3}{3}+\frac{x^5}{5}\bigg)\bigg(1-\frac{x^2}{2}+\frac{x^4}{4}\bigg)}$

But that didn't really seem to have helped =/
any ideas?

**galactus** · December 26th 2009, 10:31 AM

$\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})\frac{\sqrt{1+{\pi}x}-1}{sin(2x)}$

L'Hopital:

$\lim_{x\to 0}\frac{1}{4}\frac{\pi}{\sqrt{1+{\pi}x}\cdot cos(2x)}\cdot tan^{-1}(\frac{1}{|x|})$

The radical term can be eliminated because it evaluates to 1:

$\frac{\pi}{4}\frac{\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})}{\lim_{x\to 0}cos(2x)}$

The cos term in the denominator evaluates to 1 and we have:

$\frac{\pi}{4}\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})$

$\lim_{x\to 0}tan^{-1}(\frac{1}{|x|})=\frac{\pi}{2}$

So, we have $\frac{\pi}{4}\cdot \frac{\pi}{2}=\boxed{\frac{{\pi}^{2}}{8}}$

**tonio** · December 26th 2009, 10:48 AM

Originally Posted by Jones

Hi,

I have this limit i would like some help with

$\displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{sin(2x)}$

So i tried replacing the sin(2x) term by it's equivalent taylor polynomials for (sin x and cos x)

like so:
$\displaystyle\lim_{x\to 0}~arctan(1/|x|)~\frac{\sqrt{1+\pi x}-1}{2\bigg(\frac{x^3}{3}+\frac{x^5}{5}\bigg)\bigg(1-\frac{x^2}{2}+\frac{x^4}{4}\bigg)}$

But that didn't really seem to have helped =/
any ideas?

Perhaps there's a mistake since the limit, as given, doesn't exist: the left and right limits are different, but if we look for the limit when $x\rightarrow 0^+$ then:
your idea looks just fine to me, but instead of writing the MacClaurin series for sine and cosine let us use the series for $\sin 2x=2x+\frac{(2x)^3}{3!}+\ldots$ , and now we can try L'Hospitals rule on that fraction, since $\arctan\frac{1}{|x|}\xrightarrow [x\to 0]{}\frac{\pi}{2}$ , so derivating numerator and denominator we get:

$\lim_{x\to 0}\frac{\displaystyle{\frac{\pi}{2\sqrt{1+\pi x}}}}{2+4x^2+\ldots}=\frac{\pi}{4}$ , and thus the limit is $\frac{\pi^2}{8}$

Tonio

**Soroban** · December 26th 2009, 01:05 PM

Hello, Jones!

$\displaystyle\lim_{x\to 0}\,\arctan\!\left(\frac{1}{|x|}\right)\,\frac{\sq rt{1+\pi x}-1}{\sin2x}$

Note that: . $\arctan\left(\frac{1}{|x|}\right) \:=\:\text{arccot}|x|$

Multiply by $\frac{\sqrt{1+\pi x} + 1}{\sqrt{1+\pi x} + 1}$

. . $\text{arccot}|x|\cdot\frac{\sqrt{1+\pi x} - 1}{\sin2x}\cdot\frac{\sqrt{1+\pi x} + 1}{\sqrt{1+\pi x} + 1} \;=\;\text{arccot}|x|\cdot\frac{(1+\pi x) - 1}{\sin2x(\sqrt{1+\pi x} + 1}$

. . $=\; \text{arccot}|x|\cdot\frac{\pi x}{\sin2x(\sqrt{1+\pi x} + 1)} \;=\;\frac{\pi}{2}\cdot\text{arccot}|x|\cdot\frac{ 1}{\sqrt{1+\pi x} + 1}\cdot \frac{2x}{\sin2x}$

$\lim_{x\to0}\left[\frac{\pi}{2}\cdot\text{arccot}|x| \cdot\frac{1}{\sqrt{1+\pi x} + 1} \cdot\frac{2x}{\sin2x}\right]$

. . $=\; \frac{\pi}{2}\cdot \underbrace{\left(\lim_{x\to0}\text{arccot}|x|\rig ht)}_{\frac{\pi}{2}} \cdot \underbrace{\left(\lim_{x\to0}\frac{1}{\sqrt{1+\pi x} + 1}\right)}_{\frac{1}{2}} \cdot \underbrace{\left(\lim_{x\to0}\frac{2x}{\sin2x}\ri ght)}_{1}$

. . $=\;\frac{\pi^2}{8}$

**Jones** · December 27th 2009, 12:38 AM

Hm, difficult limit. I don't think i would have pulled this one off

Thanks for the help though

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