# [SOLVED]Limit of exponential function

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
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.
The series expansion of $\cos x = 1 - \frac 12x^2 + \ldots$.
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.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
It's like this:
$$\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)\\ =1\cdot 1 -\frac 12 x^2 \cdot 1+1\cdot \frac 12 (2x)^2 - \frac 12 x^2 \cdot \frac 12 (2x)^2 + \text{ other terms with }x^4\text{ and higher order}\\ =1 + \Big(-\frac 12 + \frac 12(2^2)\Big)x^2 + \ldots \\ = 1+\frac 32 x^2 + \ldots$$

#### goody

##### New member
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
I substituted $n=\frac 1{x^2}$, which also means that $x^2=\frac 1 n$.
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 $y=\frac 32$...

#### goody

##### New member
Thank you very much for your time and explanation. You really helped me a lot!

#### Prove It

##### Well-known member
MHB Math Helper
Re: Help with limit

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

\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.

#### goody

##### New member
It's like this:
$$\Big(1-\frac 12 x^2 + \ldots\Big)\Big(1+\frac 12 (2x)^2 - \ldots\Big)\\ =1\cdot 1 -\frac 12 x^2 \cdot 1+1\cdot \frac 12 (2x)^2 - \frac 12 x^2 \cdot \frac 12 (2x)^2 + \text{ other terms with }x^4\text{ and higher order}\\ =1 + \Big(-\frac 12 + \frac 12(2^2)\Big)x^2 + \ldots \\ = 1+\frac 32 x^2 + \ldots$$
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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?
It is correct and you are correct too.

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}

#### Prove It

##### Well-known member
MHB Math Helper
So much unneccessary analysis when L'Hospital's Rule is so concise...

#### Klaas van Aarsen

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

MHB Math Helper

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh really?
We can see the original OP in my quote in post #2: Also note in your screenshot the: Last edited by goody; April 6th, 2020 at 3:09.

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

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.

#### Prove It

##### Well-known member
MHB Math Helper
Ah well, I was kind of happy to see that the OP showed interest in power series expansions.
They are kind of... well... powerful.
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.