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[SOLVED] Limit of exponential function

goody

New member
Mar 31, 2020
14
Hello everyone, can anybody solve this limit? This is really tough one for me, thank you in advance.
limita.png
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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.
 

goody

New member
Mar 31, 2020
14
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?
limita2.png
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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
Mar 31, 2020
14
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
Mar 5, 2012
8,736
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
Mar 31, 2020
14
Thank you very much for your time and explanation. You really helped me a lot!
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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
Mar 31, 2020
14
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 1.png.
Is it correct? Because I think it should be like that 2.png or did I miss something?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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
Jan 26, 2012
1,403
So much unneccessary analysis when L'Hospital's Rule is so concise...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Oh really?
We can see the original OP in my quote in post #2:

original_OP.jpg

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
Jan 26, 2012
1,403
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.