Evaluate $\lim_{x\to\infty}\left(\frac{x^2}{2x+1}\right)^{\frac1x}$

Question

How do you evaluate the following limit? $$ \lim_{x\to\infty}\left(\frac{x^2}{2x+1}\right)^{\frac1x} $$

I tried to make it approach the shape of the euler limit

$$ \lim_{x\to\infty}\left(1+\frac{1}{f(x)}\right)^{f(x)} $$ I added and subtracted 1, so that we get "1 + a function", I inverted this function, making 1 / (1 / f (x)). then to the exponent I multiplied and divided by the function, in order to obtain a limit like that of euler, raised to another function but I realized that it was not the solution, because f (x) did not go to infinity when x went to infinity.

Then I gave up and asked here.

Welcome to Mathematics Stack Exchange, Pasqualone. I'm afraid to say that I downvoted this question because it does meet the standards of the site, as set out in this thread. — Joe, Sep 10 '21 at 20:42
To improve your question, I would suggest that you (1) add your own attempts if possible, (2) tell us which textbook this question came from, and (3) briefly tell us your mathematical background—what is your level of education, and what course are you studying? — Joe, Sep 10 '21 at 20:46
i tried the exercise several times and i couldn't, i found the exercise on youmath, i just asked for help — Pasqualone, Sep 10 '21 at 20:51
@Pasqualone That's fine but to obtain some help it is necessary that you show your work and effort on that by editing and completing the question. What have you tried exactly? — user, Sep 10 '21 at 20:52
@Pasqualone: Okay, so can you edit your question to include your attempts? It would also be wise to indicate where you got this question from. And finally, can you tell us about your mathematical background, so that the answers are pitched at the right level? — Joe, Sep 10 '21 at 20:53
As to your assessing your level - are you aware of the limit definition of $e$? L'Hopital's rule? — user170231, Sep 10 '21 at 20:55
Yes, although I'm not mistaken that the exercise wanted me to solve it without the hospital — Pasqualone, Sep 10 '21 at 20:58
@Pasqualone It is indeed solveable without l'Hospital but you need to edit your question in order to reopen it and obtain some indication or help for the solution. — user, Sep 10 '21 at 21:03
no, I added and subtracted 1, so that we get "1 + a function", I inverted this function, making 1 / (1 / f (x)). then to the exponent I multiplied and divided by the function, in order to obtain a limit like that of euler, raised to another function but I realized that it was not the solution, because f (x) did not go to infinity when x went to infinity — Pasqualone, Sep 10 '21 at 21:19
@Pasqualone Have you tried with the exponential trick $f(x)^{g(x)}=e^{g(x)\log(f(x))}$? — user, Sep 10 '21 at 21:37
@Pasqualone Well done for the editing! Can you see the difference and how much it is important give the necessary context to the question? — user, Sep 10 '21 at 21:44
@Pasqualone: Well done for improving your question! We've reopened your post, and I have removed my downvote. — Joe, Sep 10 '21 at 22:25

score 6 · Accepted Answer · answered Sep 10 '21 at 21:42

6

We can use exponential trick in this way

$$\left(\frac{x^2}{2x+1}\right)^{1/x}=e^{\frac{\log\left(\frac{x^2}{2x+1}\right)}x}\to e^0=1$$

indeed by standard limit $\frac{\log x}x \to 0$ we have

$$\frac{\log\left(\frac{x^2}{2x+1}\right)}x=\frac{\log\left(\frac{x^2}{2x+1}\right)}{\frac{x^2}{2x+1}}\frac{x}{2x+1}\to 0\cdot \frac 12 =0$$

Evaluate $\lim_{x\to\infty}\left(\frac{x^2}{2x+1}\right)^{\frac1x}$

1 Answers1