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- Mar 5, 2012

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The series expansion of $\cos x = 1 - \frac 12x^2 + \ldots$.Hello everyone, can anybody solve this limit without using L'Hospital's rule? This is really tough one for me, I know I'd use that x = e^lnx.

And the expansion of $\frac 1{1-x} = 1+x+x^2+\ldots$

So:

$$\frac{\cos(x)}{\cos(2x)}

= \frac{1-\frac 12 x^2 + \ldots}{1-\frac 12 (2x)^2 + \ldots}

=\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)

=1+\frac 32 x^2+\ldots

$$

Consequently:

$$\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}

=\lim_{x\to 0}(1+\frac 32 x^2+\ldots)^{1/x^2}

=\lim_{n\to\infty} (1+\frac {3/2}n)^n

=e^{3/2}

$$

To be fair, that last step where I leave out the dots is not entirely rigorous, but we do get the result.

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- Mar 5, 2012

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It's like this:I understand that you used Maclaurin series expansion and then the first step behind first equal sign but may I ask how did we get another steps?

$$\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)\\

=1\cdot 1 -\frac 12 x^2 \cdot 1+1\cdot \frac 12 (2x)^2 - \frac 12 x^2 \cdot \frac 12 (2x)^2 + \text{ other terms with }x^4\text{ and higher order}\\

=1 + \Big(-\frac 12 + \frac 12(2^2)\Big)x^2 + \ldots \\

= 1+\frac 32 x^2 + \ldots

$$

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I substituted $n=\frac 1{x^2}$, which also means that $x^2=\frac 1 n$.Oh of course, I got it. And in the last step you just divided 3/2 by n because n is infinite number of terms behind dots or why is it like that? Why did you not divide 1 by n as well?

So we get $\Big(1 + \frac 32 x^2\Big)^{1/x^2} = \Big(1 + \frac 32 \cdot \frac 1n\Big)^n$.

Now consider the following known limit, which is one of the equivalent definitions of the exponential function:

$$\lim_{n\to\infty} \Big(1+\frac yn\Big)^n = e^y$$

Substitute $y=\frac 32$...

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This is a $\displaystyle 1^{\infty} $ indeterminate form. It can be transformed into a form that can use L'Hospital's Rule byHello everyone, can anybody solve this limit? This is really tough one for me, thank you in advance.

$\displaystyle \begin{align*}

\lim_{x \to 0} \left\{ \left[ \frac{\cos{\left( x \right) }}{\cos{ \left( 2\,x \right) }} \right] ^{1/x^2} \right\} &= \lim_{x \to 0} \left( \mathrm{e}^{\ln{ \left\{ \left[ \frac{\cos{(x)}}{\cos{(2\,x)}} \right] ^{1/x^2} \right\} }}

\right) \\

&= \lim_{x \to 0} \left\{ \mathrm{e}^{\frac{1}{x^2}\ln{ \left[ \frac{\cos{(x)}}{\cos{(2\,x)}} \right] }} \right\} \\

&= \lim_{x \to 0} \left\{ \mathrm{e}^{\frac{\ln{\left[ \frac{\cos{(x)}}{\cos{(2\,x)}} \right]}}{x^2}} \right\} \\

&= \mathrm{e}^{ \lim_{x \to 0} \left\{\frac{\ln{\left[ \frac{\cos{(x)}}{\cos{(2\,x)}} \right] }}{x^2} \right\} }\end{align*} $

This limit is now a $\displaystyle \frac{0}{0} $ indeterminate form, so you can apply L'Hospital's Rule.

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Still, now I'm wondering how we got this .It's like this:

$$\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)\\

=1\cdot 1 -\frac 12 x^2 \cdot 1+1\cdot \frac 12 (2x)^2 - \frac 12 x^2 \cdot \frac 12 (2x)^2 + \text{ other terms with }x^4\text{ and higher order}\\

=1 + \Big(-\frac 12 + \frac 12(2^2)\Big)x^2 + \ldots \\

= 1+\frac 32 x^2 + \ldots

$$

Is it correct? Because I think it should be like that or did I miss something?

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It is correct and you are correct too.Still, now I'm wondering how we got this .

Is it correct? Because I think it should be like that or did I miss something?

Note that $\frac 1{1-x}=1+x+x^2+x^3+\ldots$.

Written with $y$ we have: $(1-y)^{-1}=1+y+y^2+y^3+\ldots$

Now substitute $y=\frac 12 (2x)^2 - \ldots$ and we get:

\begin{aligned}\Big(1-\frac 12 (2x)^2 + \ldots\Big)^{-1}

&=1+\Big(\frac 12 (2x)^2 - \ldots\Big) + \Big(\frac 12 (2x)^2 - \ldots\Big)^2 + \Big(\frac 12 (2x)^2 - \ldots\Big)^3 + \ldots \\

&=1 + \frac 12 (2x)^2 - \text{ higher order terms starting from }x^4

\end{aligned}

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The original OP asked explicitly to do it without L'Hospital's Rule.So much unneccessary analysis when L'Hospital's Rule is so concise...

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We can see the original OP in my quote in post #2:Oh really?

Also note in your screenshot the:

Ah well, I was kind of happy to see that the OP showed interest in power series expansions.

They are kind of... well...

My request to goody , please don't edit an opening post after people have answered.

Or at least not without indicating that the question was edited.

Nobody's saying they aren't, I guess I'm just someone who believes in Occam's Razor, that the simplest and most concise solution is the best one.Ah well, I was kind of happy to see that the OP showed interest in power series expansions.

They are kind of... well...powerful.