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Without the use of L'hospitals rule, solve the following: $$\lim_{x\rightarrow\infty}\left(\frac{x+2}{x+1}\right)^{x/2}$$

I'm trying to apply the limit that says $$\lim_{x\rightarrow\pm\infty} \left(1+\frac{1}{x}\right)^x = e$$

However, I'm confused as the exponent is now $x/2$ and $x$ is approaching positive infinity in the limit, and not $\pm\infty$. Also there's the rational.

Thank you in advance

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  • $\begingroup$ The $\infty$ instead of $\pm \infty$ is not a problem. You can use $(x+2)/(x+1)=1+1/(x+1)$. What would be, if $x/2$ was replaced by $x$? $\endgroup$ – k1next Nov 4 '14 at 14:33
  • $\begingroup$ If it was just x, it would be equal to $e$, right? $\endgroup$ – PurpleManiac Nov 4 '14 at 14:37
  • $\begingroup$ Exactly, and does this help: math.stackexchange.com/questions/695290/… $\endgroup$ – k1next Nov 4 '14 at 14:38
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    $\begingroup$ Thank you, I thought I couldn't move the exponent outside the limit because of the x, but I just realized I could write $x/2$ as $(expr^x)^{1/2}$ $\endgroup$ – PurpleManiac Nov 4 '14 at 14:40
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From here we take \begin{align} \lim_{n\rightarrow \infty} \sqrt{a_n} = \sqrt{\lim_{n\rightarrow \infty} a_n} \end{align}

We can rewrite your expression as: \begin{align} \lim_{x\rightarrow \infty} \Bigl( \frac{x+2}{x+1}\Bigr)^{x/2} &= \lim_{x\rightarrow \infty} \sqrt{\Bigl( \frac{x+2}{x+1}\Bigr)^{x}} = \sqrt{\lim_{x\rightarrow \infty} \Bigl( \frac{x+2}{x+1}\Bigr)^{x}} \\ &= \sqrt{\lim_{x\rightarrow \infty} \Bigl( 1+\frac{1}{x+1}\Bigr)^{x}} \\ &=\sqrt{e}\end{align} where we use the fact, that you know the limit of $(1+1/x)^x$.

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    $\begingroup$ But $\frac{1}{x + 1} \not\equiv \frac{1}{x}$ . How did you use that limit then? $\endgroup$ – Nick Nov 4 '14 at 15:03
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    $\begingroup$ You can say $\lim_{x\to\infty} \left(1+\frac{1}{x+1}\right)^x = \lim_{x\to\infty} \left(1+\frac{1}{x+1}\right)^{x+1} \times \left(1+\frac{1}{x+1}\right)^{-1} = \lim_{x\to\infty} \left(1+\frac{1}{x+1}\right)^{x+1} \times \lim_{x\to\infty} \left(1+\frac{1}{x+1}\right)^{-1} = e\cdot 1$. $\endgroup$ – JiK Nov 4 '14 at 15:07
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$$\lim_{x\to \infty}\left(\frac{x+2}{x+1}\right)^{\frac{x}{2}}$$ $$=\lim_{x\to \infty}\left(\frac{(x+1)+1}{x+1}\right)^{\frac{(x+1)-1}{2}}$$ $$=\lim_{x\to \infty}\left(1+\frac{1}{x+1}\right)^{\frac{x+1}{2}}\cdot \lim_{x\to \infty}\left(1+\frac{1}{x+1}\right)^{\frac{-1}{2}}$$ $$=\left(\lim_{x\to \infty}\left(1+\frac{1}{x+1}\right)^{x+1}\right)^{1/2}\cdot \lim_{x\to \infty}\left(1+\frac{1}{x+1}\right)^{\frac{-1}{2}}$$ $$=\left(e\right)^{1/2}\cdot \left(1+0\right)^{\frac{-1}{2}}=\color{red}{e^{1/2}}$$

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    $\begingroup$ Why the downvote? This answer is correct (unlike at least one other) and clear. $\endgroup$ – Simon S Nov 6 '15 at 19:21
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$$\lim_{x\to\infty} (1+\frac1{x+1})^{\frac{x}2} =\lim_{x\to\infty} \left[(1+\frac1{x+1})^{\frac1{x+1}}\right]^{\frac{x}{2(x+1)}} = e^{\lim\limits_{x\to \infty} \frac{x}{2(x+1)}}$$

The last line is true because $\lim\limits_{x\to a} f(x)=f(\lim\limits_{x\to a} x)$ is true when $f$ is continuous.

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  • $\begingroup$ Not convinced by the last equality. $\endgroup$ – Lost1 Nov 4 '14 at 14:37
  • $\begingroup$ @Lost1 See the edit $\endgroup$ – Sayan Nov 4 '14 at 14:39
  • $\begingroup$ still not convinced, you are saying $\lim a^b = (\lim a)^{(\lim b)}$. Your supposedly justification does not do the job. $\endgroup$ – Lost1 Nov 5 '14 at 8:54
  • $\begingroup$ @Lost1 Yes it does. Suppose $\lim_{x\to a} f(x)=1$ and $\lim_{x\to a} g(x)=\infty$. Then convince yourself that $\lim_{x\to a} f(x)^{\frac{1}{f(x)-1}} = e$. $$\begin{aligned}\lim_{x\to a} f(x)^{g(x)} & = \lim_{x\to a} \left(f(x)^{\frac1{f(x)-1}}\right)^{g(x)(f(x)-1)}\\ & = \lim_{x\to a} e^{g(x)(f(x)-1)\ln \left(f(x)^{\frac1{f(x)-1}}\right)} \\ & = e^{\lim_{x\to a} g(x)(f(x)-1)\ln\left( \lim_{x\to a} f(x)^{\frac1{f(x)-1}}\right) } \\ & = e^{\lim_{x\to a} g(x)(f(x)-1)} \end{aligned}$$ The above stated result was applied to $e^x$ and $\ln x$ functions and $\lim ab=\lim a\lim b$ was used. $\endgroup$ – Sayan Nov 5 '14 at 10:20
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    $\begingroup$ Jumped way too many steps with the last equality, even with the comment. Given that the OP asked this question, I am pretty convinced he could not have filled all these gaps himself. $\endgroup$ – Lost1 Nov 5 '14 at 13:37
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$$\lim_{x\to\infty}(\frac{x+2}{x+1})^{\frac{x}{2}}=\lim_{x\to\infty}(1+\frac{1}{x+1})^{\frac{x}{2}}$$ Consider now the function $$f(y)=\frac{\ln(1+y)}{\frac{2y}{1-y}}=(1-y)\frac{\ln(1+y)}{2y}$$ for all $y>0$ and $f(0)=\frac{1}{2}$ for $y=0$. Notice that $$\lim_{y\to0}f(y)=\frac{1}{2}$$ implying that $f(y)$ is continous at $y=0$. Set $y=\frac{1}{x+1}$ then $$\lim_{y\to0}f(y)=\lim_{y\to0}\{(1-y)\frac{\ln(1+y)}{2y}\}=\lim_{x\to\infty}\{\frac{x}{2}\ln(1+\frac{1}{1+x})\}$$ But $$\lim_{y\to0}f(y)=\frac{1}{2}$$ Therefore $$\lim_{x\to\infty}\{\frac{x}{2}\ln(1+\frac{1}{1+x})\}=\frac{1}{2}$$ Since $$e^{\frac{x}{2}\ln(1+\frac{1}{1+x})}=(1+\frac{1}{1+x})^{\frac{x}{2}}$$ then $$\lim_{x\to\infty}(1+\frac{1}{1+x})^{\frac{x}{2}}=e^{1/2}$$

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Consider $$A=\left(\frac{x+2}{x+1}\right)^{\frac x2}$$ and take logarithms of both sides. $$\log(A)=\frac x2\, \log\left(\frac{x+2}{x+1}\right)=\frac x2 \,\log\left(1+\frac{1}{x+1}\right)$$ Now, remembering that $\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$ and making $\epsilon =\frac{1}{x+1}$ , we then have $$\log(A)=\frac x2\left(\frac{1}{x}-\frac{3}{2 x^2}+O\left(\frac{1}{x^3}\right)\right)=\frac{1}{2}-\frac{3}{4 x}+O\left(\frac{1}{x^2}\right)$$ and Taylor again $$A=e^{\log(A)}=\sqrt e\left(1+\frac{3}{4 x}\right)+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.

Numerically, the expansion gives a relative error smaller than $1$% as soon as $x>12$ and smaller than $0.1$% as soon as $x>37$.

For illustration purposes, for $x=50$, the function value is $\approx 1.62491$ while the above asymptotics gives $\approx 1.623991$.

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