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Find the limit $\displaystyle \lim_{x \to0^+} (\sin x)^{\large {\frac1 {\ln x}}}$

Here's what I did:

$\begin{align}\lim_{x \to 0^+} (\sin x)^{\large \frac1{\ln x}}&(1) =\lim e^{\large {\ln( { (\sin x)^\frac1{\ln x}})}} \ \ \\ &(2) = \lim_{x \to0^+} e^{\large {{\frac1{\ln x}}\ln( { (\sin x)})}} \ \ \\ & (3)= \lim_{x \to0^+} e^{\large {\frac{\ln\sin x}{\ln x}}} \ \ \\ &(4) = \lim_{x \to0^+} e^{\large {\ln\frac{ \sin x}{ x}}} \ \\ & (5)= \lim_{x \to0^+} \frac{ \sin x}{ x} = 1 \end{align}$

But I know this is wrong because when I plot it it goes to e.

I am now stuck on step 3.

Any advice please ?

PS: we can't use derivation or integration, nor Taylor's theorem because we haven't covered that.

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4 Answers 4

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Your mistake is in the fourth equality: It is not true that

$$\ln \left(\frac a b\right) = \frac{\ln a}{\ln b}$$

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METHOD 1: l'Hopital's Rule

I will add to the answer provided by T. Bongers:

So, you have: $$\lim \exp(\frac{\ln(\sin(x))}{\ln(x)}) \\ = \lim \exp(\frac{\cos(x)/\sin(x)}{1/x}) \\ = \lim \exp(\frac{\cos(x)\cdot x}{\sin(x)}) \\ = \exp(\lim [(\frac{\sin(x)}{x})^{-1}\cdot \cos(x)]) =\exp(1).$$

METHOD 2: Squeeze Theorem and Taylor's Theorem

As in method 1, we have $$\lim \exp(\frac{\ln(\sin(x))}{\ln(x)}) \\ \leq \lim \exp(\frac{\ln(x)}{\ln(x)}) = e.$$

Also, $$\lim \exp(\frac{\ln(\sin(x))}{\ln(x)}) \\ \geq \lim \exp(\frac{\ln(x)+\ln(1-x^2/6)}{\ln(x)}) \\ \geq e\cdot \lim \exp(\frac{-x^2/6-x^4/(72(1+(x^2/6))^2)}{x}) \\ \geq e\cdot \exp( \lim (-x/6-x^3/(72(1+(x^2/6))^2))) = e\cdot 1 = e.$$

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    $\begingroup$ I never saw this exp notation... $\endgroup$
    – GinKin
    Commented Dec 13, 2013 at 21:21
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    $\begingroup$ Well, $\exp(x) = e^x$. Nothing new here. $\endgroup$ Commented Dec 13, 2013 at 21:21
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    $\begingroup$ OK, how can you break the ln into these fractions ? $\endgroup$
    – GinKin
    Commented Dec 13, 2013 at 21:22
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    $\begingroup$ You can't. Here you want to use l'Hopital's Rule. $\endgroup$ Commented Dec 13, 2013 at 21:23
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    $\begingroup$ Oh I see. We'll we can't use derivations yet because we haven't covered that. $\endgroup$
    – GinKin
    Commented Dec 13, 2013 at 21:24
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continuing from step 3 we have $$\displaystyle = \lim_{x \to0^+} e^{\large {\frac{\ln \sin x}{\ln x}}}$$ $$=\displaystyle e^{\lim_{x \to0^+} {\large {\frac{\ln \sin x}{\ln x}}}}$$

So the problem boils down to evaluating $$=\displaystyle \lim_{x \to0^+} {\large {\frac{\ln \sin x}{\ln x}}}$$

I claim that for $0<x<\pi/2$ that the following holds $$\sin x \lt x \lt \tan x$$ Figure 1
In figure 1, we let $OC=OA=1$. In other words, $Arc\:CA=x$ is an arc of a unit circle. The shortest distance from point $C$ to line $AO$ is line $CE=\sin x$ (because $CE\perp OA$). Another path from point $C$ to line $OA$ is arc $CA$ (which is longer than CE because it is not the shortest path). So we have at the very least $$\sin x \lt x$$ Now we need to show that line $BA=\tan x \gt x$.
Lines $AD$ and $CD$ are both tangent to arc $CA$. $CD+DA$ is longer than arc $CA$ because the set of points bound by sector $OCA \subset $the set of points bound by quadrilateral $OCDA$, both of which are convex sets. This means that the perimeter of quadrilateral $OCDA$ must be longer than the perimeter of the sector $OCA$ (as per Archimedes, On the Sphere and Cylinder Book I). But both the sector and the quadrilateral both have sides $OC$ and $OA$, so we have $$CA=x<DC+DA$$ But $BD>CD$ because it is the hypotenuse in $\triangle BCD$ we have $$\tan x = BA = BD+DA\gt CD+DA \gt CA=x \gt \sin x$$

Because $\sin x$ is closer to zero than $x$ which is closer to zero than $\tan x$, $\ln \sin x$ must be closer to $-\infty$ than $\ln x$ which must be closer to $-\infty$ than $\ln\tan x$, so we have $$\ln\sin x \lt \ln x \lt \ln\tan x$$ $$\ln\sin x \lt \ln x \lt \ln\frac{\sin x}{\cos x}$$ $$\ln\sin x \lt \ln x \lt \ln\sin x - \ln\cos x$$ $$\frac{\ln\sin x}{\ln x} \gt 1 \gt \frac{\ln\sin x}{\ln x} - \frac{\ln\cos x}{\ln x}$$ or $$\frac{\ln\sin x}{\ln x} - \frac{\ln\cos x}{\ln x}\lt 1 \lt \frac{\ln\sin x}{\ln x} $$ But $$\lim_{x \to0^+}\frac{\ln\cos x}{\ln x}=0$$ so we have $$\lim_{x \to0^+}\frac{\ln\sin x}{\ln x} \le 1 \le \lim_{x \to0^+}\frac{\ln\sin x}{\ln x} $$ which implies that $$\lim_{x \to0^+}\frac{\ln\sin x}{\ln x} = 1$$ thus $$\displaystyle \lim_{x \to0^+}\bigg(\sin x\bigg)^{\frac{\displaystyle 1}{\displaystyle\ln x}} = e$$

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  • $\begingroup$ Very nice answer. $\endgroup$
    – GinKin
    Commented Apr 30, 2014 at 17:17
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Without using hopital or calculus we can do it this way:

For x we have $\sin(x) \le x$ and so therefore $\ln(\sin(x)) \le \ln(x)$ therefore: $(\sin(x))^{1/\ln(x)} \le e$

but $\frac{\sin(x)}{x} \le \frac{\ln(\sin(x))}{\ln(x)} $ because $a/b \le \log_b(a)$

therefore $e^{\frac{\sin(x)}{x}} \le (\sin(x))^{\frac{1}{\ln(x)}} \le e$

and now by the squeeze theorem $\lim_{x \to 0^+} (\sin(x))^{\frac{1}{\ln(x)}} = e$

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  • $\begingroup$ Why is that true: $(\sin(x))^{1/\ln(x)} \le e$ ? $\endgroup$
    – GinKin
    Commented Dec 14, 2013 at 21:55
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    $\begingroup$ $\ln(\sin(x))/\ln(x) \le 1$ therefore $(\sin(x))^{1/\ln(x)} = e^{\ln(\sin(x))/\ln(x)} \le e^1 = e$ $\endgroup$
    – JmsNxn
    Commented Dec 14, 2013 at 21:59
  • $\begingroup$ $a/b \le \log_b(a)$ isn't true. $\ln x > x/e$ ? $\endgroup$
    – GinKin
    Commented Dec 23, 2013 at 19:44
  • $\begingroup$ $-5<-3,$ but $\frac{-5}{-3}>1$. $\endgroup$
    – PinkyWay
    Commented Oct 28, 2020 at 14:28

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