limits ( the indeterminate forms)

Oct 2018
33
3
PH
Hi guys, anyone can help me about this topic.

Thanks a lot.


Sent from my SM-J730G using Tapatalk
 
Dec 2011
2,314
916
St. Augustine, FL.
Is it:

\(\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)\) ?
 
  • Like
Reactions: 1 person
Oct 2018
33
3
PH
Is it:

\(\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)\) ?
yes thats exactly the problem.

Sent from my SM-J730G using Tapatalk
 
Dec 2011
2,314
916
St. Augustine, FL.
yes thats exactly the problem.
Okay, what I would do is set:

\(\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L\)

And take the natural log of both sides:

\(\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)\)

Now, because log functions are one-to-one, we can bring it inside the limit:

\(\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)\)

Now, applying a log property, we may write:

\(\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)\)

Now, we have the indeterminate form \(\displaystyle \frac{\infty}{\infty}\) on the LHS and can apply L'Hôpital's Rule...what do you get?
 
  • Like
Reactions: 1 person
Oct 2018
33
3
PH
Okay, what I would do is set:

\(\displaystyle \lim_{x\to\infty}\left(\left(1+x^2e^x\right) ^{\Large\frac{1}{x}}\right)=L\)

And take the natural log of both sides:

\(\displaystyle \ln\left(\lim_{x \to\infty}\left( \left(1+x^2e^x \right) ^{\Large\frac{1}{x}}\right)\right) =\ln(L)\)

Now, because log functions are one-to-one, we can bring it inside the limit:

\(\displaystyle \lim_{x\to\infty} \left(\ln\left( \left(1+x^2e^x \right) ^{\Large \frac{1}{x}}\right) \right) =\ln(L)\)

Now, applying a log property, we may write:

\(\displaystyle \lim_{x\to\infty} \left(\frac{\ln \left(1+x^2e^x\right)}{x} \right) =\ln(L)\)

Now, we have the indeterminate form \(\displaystyle \frac{\infty}{\infty}\) on the LHS and can apply L'Hôpital's Rule...what do you get?
thanks wait i will rewrite this equation and see if i can get this.

Sent from my SM-J730G using Tapatalk
 
Dec 2011
2,314
916
St. Augustine, FL.
To follow up, using L'Hôpital's Rule, we may write:

\(\displaystyle \lim_{x\to\infty} \left(\frac{\dfrac{2xe^x+x^2e^x}{1+x^2e^x}}{1} \right) =\ln(L)\)

Or:

\(\displaystyle \lim_{x\to\infty} \left(\frac{x^2+2x}{e^{-x}+x^2} \right) =\ln(L)\)

Applying L'Hôpital's Rule again, we get:

\(\displaystyle \lim_{x\to\infty} \left(\frac{2x+2}{2x-e^{-x}} \right) =\ln(L)\)

Or:

\(\displaystyle \lim_{x\to\infty} \left(\frac{2+ \dfrac{2} {x}}{2- \dfrac{1}{xe^{x}}} \right) =\ln(L)\)

Hence:

\(\displaystyle 1=\ln(L)\implies L=e\)
 
Dec 2016
302
165
Earth
You can avoid L'Hôpital's Rule if you would show that


\(\displaystyle \displaystyle\lim_{x \to \infty}(e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^\sqrt{x}e^x)^{1/x} \ \ \ \ \)**


\(\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{x + \sqrt{x}})^{1/x}\)


\(\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ \displaystyle\lim_{x \to \infty}(e^{1 + \tfrac{1}{\sqrt{x}}}) \)


\(\displaystyle e \ \le \ \displaystyle\lim_{x \to \infty}(1 + x^2e^x)^{1/x} \ \le \ e \)




** You would want to make sure that you can show that \(\displaystyle \ \ \displaystyle\lim_{x \to \infty}\bigg(\dfrac{e^\sqrt{x}e^x}{1 + x^2e^x}\bigg) > 1\).
 
Last edited: