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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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: $$\lim\limits_{x\rightarrow\infty}u^v=\lim_{x\rightarrow\infty}e^{v\ln{u}}=e^{\lim\limits_{x\rightarrow\infty}v\ln{u}}=e^{\lim\limits_{x\rightarrow\infty}v\ln\lim\limits_{x\rightarrow\infty}u}=\left(\lim_{x\rightarrow\infty}u\right)^{\lim\limits_{x\rightarrow\infty}v}.$$

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