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I'm struggling a bit with solving a limit problem using L'Hopital's Rule:

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

My work:

$$y = \left(1+\frac{1}{x}\right)^{2x}$$ $$\ln y = \ln \left(1+\frac{1}{x}\right)^{2x} = 2x \ln \left(1+\frac{1}{x}\right)$$ $$=\frac{\ln \left(1+\frac{1}{x}\right)}{(2x)^{-1}}$$ Taking derivatives of both the numerator and the denominator: $$f(x) = \ln\left(1+\frac{1}{x}\right)$$ $$f'(x) = \left(\frac{1}{1+\frac{1}{x}}\right)\left(-\frac{1}{x^2}\right) = (x+1)\left(-\frac{1}{x^2}\right) = -\frac{(x+1)}{x^2}$$ $$g(x) = (2x)^{-1}$$ $$g'(x) = (-1)(2x)^{-2}(2) = -\frac{2}{(2x)^{2}} = -\frac{1}{2x^{2}}$$

Implementing the derivatives:

$$\lim_{x\to\infty} \frac{-\frac{(x+1)}{x^2}}{-\frac{1}{2x^2}} = -\frac{(x+1)}{x^2} \cdot \left(-\frac{2x^2}{1}\right) = 2(x+1) = 2x+2$$

However, I'm not sure where to go from here. If I evaluate the limit, it still comes out to infinity plus 2, and I don't know how much further to take the derivative or apply L'Hopital's Rule.

Any suggestions would be appreciated!

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  • $\begingroup$ It seems that $f'(x)$ is wrong. $f'(x) = (\frac{1}{1+\frac{1}{x}})(-\frac{1}{x^2})=-\frac{1}{x(x+1)}$ $\endgroup$ – Claude Leibovici Mar 25 '15 at 8:32
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$$2\lim_{x\to\infty}\frac{\ln(1+x)-\ln x}{(x)^{-1}}$$

$$=2\lim_{x\to\infty}\frac{1/(1+x)-1/x}{-1/x^2}$$

$$=2\lim_{x\to\infty}\frac{x^2}{x(1+x)}$$

$$=2\lim_{x\to\infty}\frac1{1/x+1}=\frac2{0+1}$$

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  • $\begingroup$ What am I missing here? The answer to the question is clearly $e^2$, right? Are you answering some other part of the question? $\endgroup$ – barak manos Mar 25 '15 at 8:14
  • $\begingroup$ @barakmanos, Read the question completely. He actually continued the work of OP and found the value of $\ln(L)$ where $L$ is the limit. The answer is indeed $e^2$. $\endgroup$ – Prasun Biswas Mar 25 '15 at 8:15
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    $\begingroup$ There's an elementary limit identity: $\displaystyle\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e$. That trivializes the problem and solves it without using L'Hopital's rule. $\endgroup$ – Prasun Biswas Mar 25 '15 at 8:16
  • $\begingroup$ But I believe I'm required to solve using L'Hopital's Rule, as the identity isn't discussed in the coursework. Is this possible? $\endgroup$ – thisisanon Mar 25 '15 at 8:22
  • $\begingroup$ @ptikobj You can prove the identity using LHR. See my answer if you want. $\endgroup$ – user198044 Mar 25 '15 at 8:47
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$\lim_{x \to \infty} (1+1/x)^{2x}$

$= \lim_{x \to \infty} \exp(\ln(1+1/x)^{2x})$

$= \lim_{x \to \infty} \exp(2x \ln(1+1/x))$

$= \lim_{x \to \infty} \exp(2 \frac{\ln(1+1/x)}{1/x})$

$= \exp(2 \lim_{x \to \infty} \frac{\ln(1+1/x)}{1/x}) \because e^{2x}$ is continuous.

Using LHR, we get

$= \exp(2 \lim_{x \to \infty} \frac{\frac{1}{(1+1/x)} (-1/x^2)}{-1/x^2})$

$=e^2$

Btw, I think you should know from PreCalculus (in a non-rigorous sense) that

$\lim_{x \to \infty} (1+1/x)^{x} = e$

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