Exponential limits

P

Philip K.

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Apr 28, 2020
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Hello, i have weird types of limits to solve, for which i cannot find information online.
For example:
lim x->inf [(x-2)/(x+1)]^4x+1
or
lim x->inf [(x^2-2x+1)/(x^2-4x+3)]^x-2

I have the solution of the first one, but unfortunately nothing is explained and honesly I have no idea where half of the tranformations come from. Attached is a picture of the solution i have.

Can someone, please, explain it in detail? Alternative solution works as well.
 

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You got the correct answer. Good job! I would have immediately used logs to do this limit. I think that you lucked out using your method because the answer involved e (which it usually does, but not always I suspect).

Go ahead and try doing the problem again using logs. Call the limit L and then take logs of both sides. Please post back.


Also please put parenthesis around your exponents!!!!
 
Hello, and welcome to FMH! :)

Let's take a look at the first one, for which I would write:

[MATH]L=\lim_{x\to\infty}\left(\left(\frac{x-2}{x+1}\right)^{4x+1}\right)[/MATH]
Let's take the natural log of both sides as follows:

[MATH]\ln(L)=\lim_{x\to\infty}\left(\ln\left(\left(\dfrac{x-2}{x+1}\right)^{4x+1}\right)\right)[/MATH]
[MATH]\ln(L)=\lim_{x\to\infty}\left(\frac{\ln\left(\dfrac{x-2}{x+1}\right)}{\dfrac{1}{4x+1}}\right)[/MATH]
Now we have the indeterminate form 0/0 and may apply L'Hôpital's Rule:

[MATH]\ln(L)=\lim_{x\to\infty}\left(\frac{\dfrac{3}{x^2-x-2}}{-\dfrac{4}{(4x+1)^2}}\right)=-\frac{3}{4}\lim_{x\to\infty}\left(\frac{(4x+1)^2}{x^2-x-2}\right)=-12[/MATH]
Hence:

[MATH]L=e^{-12}[/MATH]
 
Thank you very much. I have 7,8 more to try. I will try them and write if i need further help. Once more, thank you !
 
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