My question is about the procedure for this limit problem: $$\lim\limits_{x \to { \infty } } (\frac{x}{x+2})^x$$

My solution was like that:

$$(\frac{x}{x+2})^x=e^{x\ln\frac{x}{x+2}} = e^u$$ with $\ u = x \ln(\frac{x}{x+2})$.

Then $$\lim\limits_{x \to { \infty } } x\ln(\frac{x}{x+2}) =\lim\limits_{x \to { \infty } }{\ln{x\over x+2}\over {1\over x}}$$

Applying L'Hôpital's rule:

$$\lim\limits_{x \to { \infty } } -{{2\over x(x+2)}\over {1\over x^2}} = \lim\limits_{x \to { \infty } } -{2x\over x+2} = -2 $$

$$\lim\limits_{x \to { \infty } } u = -2 $$

$\lim\limits_{u \to { \ -2 } } e^{u} = e^{-2} = {1\over e^{2}} $

However, according to my answer sheet, the correct answer is $e^{2}$. So, Please I need to know where's my mistake here.

Thank you.

  • 7
    $\begingroup$ The answer given to you is wrong; note that $\left(\frac{x}{x + 2}\right)^x$ is never larger than $1$, hence its limit cannot be larger than $1$. $\endgroup$ – Theo Bendit Jul 26 '19 at 2:20
  • 4
    $\begingroup$ Your mistake is in doubting your work and trusting the book to have no typos. $e^{-2}$ is correct. $\endgroup$ – Graham Kemp Jul 26 '19 at 2:38

Your answer is correct.

Also, we have $$\lim_{x\rightarrow\infty}\left(\frac{x}{x+2}\right)^x=\lim_{x\rightarrow\infty}\left(1-\frac{2}{x+2}\right)^{-\frac{x+2}{2}\cdot\frac{-2x}{x+2}}=e^{-2}.$$ I used the following property.

Let there is $\lim\limits_{x\rightarrow\infty}u(x)>0$, $\lim\limits_{x\rightarrow\infty}u(x)\neq1$ and there is $\lim\limits_{x\rightarrow\infty}v(x).$

Thus, since $e^x$ and $\ln$ are continuous functions, we obtain: $$\lim\limits_{x\rightarrow\infty}u^v=\lim_{x\rightarrow\infty}e^{v\ln{u}}=e^{\lim\limits_{x\rightarrow\infty}v\ln{u}}=e^{\lim\limits_{x\rightarrow\infty}v\ln\lim\limits_{x\rightarrow\infty}u}=\left(\lim_{x\rightarrow\infty}u\right)^{\lim\limits_{x\rightarrow\infty}v}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.