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I have to find the limit of $(\sqrt{4+x}-1)^{1/(e^x-1)}$ as $x\to0$ without de l'Hopital's rule and only with notable limits

amWhy
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Dipok
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4 Answers4

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For $x\to 0, \operatorname{e}^x\sim 1+x$ and $\sqrt{4+x}=2\left(1+\frac{x}{4}\right)^{\frac{1}{2}}\sim 2\left(1+\frac{x}{8}\right)$ so that $$ \left(\sqrt{4+x}-1\right)^{1}{\operatorname{e}^x-1}\sim \left(1+\frac{x}{4}\right)^{\frac{1}{x}} $$ Putting $y=\frac{1}{x}$ the limit is $$ \lim_{y\to\infty}\left(1+\frac{1/4}{y}\right)^y=\operatorname{e}^{1/4} $$

meta_warrior
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alexjo
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    Isn't $;e^x\sim 1+x;$ based on the Taylor expansion of the exponential function and, thus, on derivatives? The same for the rest of the evaluations for $;x;$ close to zero. I'm not sure this is allowed if l'Hospital (and apparently also derivatives) is out... – DonAntonio Nov 15 '13 at 17:55
  • It comes from the fundamental limit $\lim_x\to 0 \frac{e^x-1}{x}=1$. – alexjo Nov 15 '13 at 18:02
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    That "fundamental limit" is explained basically , as far as I know, only by seeing it equals $;e^{0}{'}=1;$ and/or passing to the limit via the inverse function $;\log x;$ ... – DonAntonio Nov 15 '13 at 18:04
  • $$\lim_{h\to 0}\frac{e^h-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\left(1+\frac{h}{n}\right)^n-1}{h}=\lim_{h\to 0}\frac{\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\left(\frac{h}{n}\right)^k-1}{h}=$$$$=\lim_{h\to 0}\lim_{n\to\infty}\frac{\sum_{k=1}^n\binom{n}{k}\left(\frac{h}{n}\right)^k}{h}=1+\lim_{h\to 0}\lim_{n\to\infty}\rlap{/}{h}\frac{\sum_{k=2}^n\binom{n}{k}\frac{h^{k-1}}{n^k}}{\rlap{/}{h}}=1$$as every single summand in what's left in that sum is multiplied by a positive power of $,h$. – alexjo Nov 15 '13 at 18:22
  • Ok, that looks better though there's still a question about taking the limit of infinite summands (this can easily be justified by remarking that sum converges, of course)...nice. +1 – DonAntonio Nov 15 '13 at 18:25
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Taking the logarithm you can compute $$ \lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{e^x-1}= \lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{x}\frac{x}{e^x-1} $$ The limit of the second fraction is $1$, so you're reduced to computing $$ \lim_{x\to 0}\frac{\log(\sqrt{4+x}-1)}{x} $$

Set $1+u=\sqrt{4+x}-1$, so $\sqrt{4+x}=u+2$ and $4+x=u^2+4u+4$, hence $x=u^2+4u$; when $x\to0$ also $u\to0$, so the limit becomes $$ \lim_{u\to 0}\frac{\log(1+u)}{u(u+4)} $$

Can you go from here?

Note: in the computation the two fundamental limits

\begin{gather} \lim_{x\to 0}\frac{e^x-1}{x}=1\\[1ex] \lim_{x\to 0}\frac{\log(1+x)}{x}=1 \end{gather}

are used. Check if you are allowed to use them.

egreg
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  • How do you know the limit of the second fraction is one? Isn't this based on the derivative of the exponential at zero? And then you're reduced to computing...etc. if you can prove both limits exist separatedly. And is the very last limit in your answer doable without derivatives/l'Hospital/Taylor...? – DonAntonio Nov 15 '13 at 17:49
  • The exponential limit $\lim_{x \to 0}\dfrac{e^{x} - 1}{x} = 1$ and logarithmic limit $\lim_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$ can be established without use of L'Hospital provided we choose any of various definitions of exponential and logarithmic function. Also since both functions are inverse to each other only one of the limits needs to be established. – Paramanand Singh Nov 16 '13 at 04:17
  • Using one complicated definition $e^{x} = \lim_{n \to \infty}(1 + (x/n))^{n}$ the exponential limit is established at http://math.stackexchange.com/questions/540816/the-derivative-of-ex-using-the-definition-of-derivative-as-a-limit-and-the-de/ Using simpler definitions like $\log x = \int_{1}^{x},dt/t$ and $e^{x} = y$ if $\ln y = x$ both the limits are established instantly without much effort. – Paramanand Singh Nov 16 '13 at 04:24
  • @DonAntonio If using l'Hôpital's theorem is not allowed, then at least the “fundamental limits” should be known; among them there are surely the ones that allow computing the derivative of the exponential and logarithmic functions. How they are obtained is another matter and depends on the teacher or the textbook. – egreg Nov 16 '13 at 10:39
  • I know that, @egreg, yet in order to avoid a circular thing here one should, imo, do something like what alexjo did in his answer... – DonAntonio Nov 16 '13 at 10:49
  • @DonAntonio I don't agree: those limits are (or should be) known to anybody doing this kind of exercises. I don't agree with prescribing not to use l'Hôpital's theorem either. – egreg Nov 16 '13 at 10:52
  • I agree with the l'H thing, @egreg, yet I can't agree with the first part: at undergraduate level one must expect to be explained things, and either one uses methods like the one used by alexjo in his answer's comments, or else wait until derivative and stuff is studied to get the explanation. I can't accept students doing something they can't explain just to be able to solve this or that problem. – DonAntonio Nov 16 '13 at 11:23
  • @DonAntonio How do you know what the OP knows at this point? I'll add something to my answer, but I find this discussion pointless. – egreg Nov 16 '13 at 11:34
  • Oh, I don't know, @egreg. I'm just basing my comments on what the OP asked : without l'H and using only notable limits (which, btw, could be precisely what you and others used! We can't know since the OP hasn't said what are those "notable limits") – DonAntonio Nov 16 '13 at 11:36
  • l'Hospital $\neq$ using derivatives, so @egreg solution is valid (in the sense that we can use Taylor theorem (for example, to prove those limits)). End of story. – Cortizol Nov 16 '13 at 11:42
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I'm not totally sure where this is heading, but here's a possible first step. $$ L = \lim_{x\to0} (\sqrt{4+x}-1)^{\large \frac{1}{e^x-1}} $$ Multiplying both sides, we have $$ \begin{align} \left[\lim_{x \to 0} (\sqrt{4+x}+1)^{\large \frac{1}{e^x-1}}\right] L &= \lim_{x\to0} [(\sqrt{4+x}-1)(\sqrt{4+x}+1)]^{\large \frac{1}{e^x-1}}\\ \left[\lim_{x \to 0} (\sqrt{4+x}+1)^{\large \frac{1}{e^x-1}}\right] \cdot L&= \lim_{x\to 0}[3+x]^{\large \frac{1}{e^x-1}} \end{align} $$ Hopefully that helps.

Ben Grossmann
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$\ln \left( \sqrt{4+x}-1 \right)^{\frac{1}{e^x-1}}= \frac{x}{e^x-1} \frac{\ln \left(1+(\sqrt{4+x}-2)) \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{\frac{x}{2+\sqrt{4+x}}} \frac{1}{2+\sqrt{4+x}} \rightarrow 1 \cdot 1 \cdot \frac{1}{4}$

Therefore $\lim_{x\to 0} \left( \sqrt{4+x}-1\right)^\frac{1}{e^x-1} = e^\frac{1}{4}$

I used well known limits $\lim_{x\to 0} \frac{e^x-1}{x}=1=\lim_{x\to 0} \frac{\ln (1+x)}{x}$

3dok
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    The "well-known" limits are based either on derivatives or l'Hospital. Can they be carried on without any of these? – DonAntonio Nov 15 '13 at 17:53
  • I've beed taught oto define $e^x= \sum_{k=0}^\infty \frac{x^k}{k!}$ ,so I don't need Hospital or derivatives. – 3dok Nov 15 '13 at 22:12
  • Nice @3dok, yet you still need to justify why $$\lim_{x\to 0}\sum_{k=1}^\infty\frac{x^k}{k!}=0$$ This is not trivial... – DonAntonio Nov 15 '13 at 22:21