What is the answer to the limit? $$\lim_{x \to 0} \left( 1 + \sin \left( \frac 3 x \right) \right)^x$$ The book's answer shows that $e^3$ while I keep getting $e^0$. I used the estimation $$\lim_{x \to 0} x \ln x \le \lim_{x \to 0} x \ln \left( 1 + \sin \left( \frac 3 x \right) \right) \le 0$$
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3Should the limit be as $x\to\infty$? As stated, the limit is $1$. – Jason Jan 27 '16 at 19:05
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No, this stupid limit is $x\to 0$ – Mula Ko Saag Jan 27 '16 at 19:06
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3That must be a typo. The limit is $e^3$ as $x \to \infty$. – Brian Tung Jan 27 '16 at 19:07
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$\sin \frac 3x$ does not converge to $+\infty$ as $x\to 0$ – user26977 Jan 27 '16 at 19:07
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@BrianTung probably – Mula Ko Saag Jan 27 '16 at 19:07
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1You need $x\to 0+$, because when $\sin(3/x)=-1$ you get $0^{x}$, which is undefined when $x<0$. – Thomas Andrews Jan 27 '16 at 19:09
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Perhaps it's $3/\sin(x)$ instead of $\sin(3/x)$? – leonbloy Jan 27 '16 at 19:10
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But it seems like it is likely undefined, since there is always a zero value and always a value close to $1$. – Thomas Andrews Jan 27 '16 at 19:11
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@leonbloy no, probably this is a typo. – Mula Ko Saag Jan 27 '16 at 19:11
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but we have $x$ tends to zero, or not? – Dr. Sonnhard Graubner Jan 27 '16 at 19:16
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1@ThomasAndrews Mathematica evaluates the limit as $1$. – Mula Ko Saag Jan 27 '16 at 19:17
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for $x$ tends to zero we get $1$ as the searched limit – Dr. Sonnhard Graubner Jan 27 '16 at 19:17
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2The limit doesn't exist since the functions takes on values of $0$ and $1$ in all $(0,\epsilon)$. – A.S. Jan 27 '16 at 19:18
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@Dr.SonnhardGraubner could you tell me more "searched limit is"? – Mula Ko Saag Jan 27 '16 at 19:19
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@A.S. thanks I understand your logic and convinced!! Perhaps you can add it as answer. – Mula Ko Saag Jan 27 '16 at 19:21
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do we have $x$ tends to $0$ or $x$ tends to $\infty$? – Dr. Sonnhard Graubner Jan 27 '16 at 19:22
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@Dr.SonnhardGraubner $x\to0$ as typo on book. Still I was having hard time believing that limit does not exist for above question. – Mula Ko Saag Jan 27 '16 at 19:23
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yes $e^3$ is the right answer – Dr. Sonnhard Graubner Jan 27 '16 at 19:24
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Good for mathematica. Doesn't make it right @MulaKoSaag – Thomas Andrews Jan 27 '16 at 19:26
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@ThomasAndrews haha yes. I was deceived. – Mula Ko Saag Jan 27 '16 at 19:27
2 Answers
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The short answer is there's a typo in the question. As $x\to\infty$, we have $$\left(1+\sin\left(\frac3x\right)\right)^x\to\exp\left(\lim_{x\to\infty}x\sin\left(\frac3x\right)\right)=e^3.$$ As pointed out in the comments, the limit as $x\to0$ does not exist.
Jason
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Pick a fixed $\alpha\geq 0$. Let $a=e^{-2\pi\alpha/3}$.
Pick $u_n\in(0,\pi/2)$ so that $\sin u_n =1-a^n$.
Then define $x_n=\frac{3}{2n\pi - u_n}$.
Then $1+\sin(3/x_n)=a^n$, and $(a^n)^x=e^{-\frac{2\pi n\alpha}{2n\pi+u_n}}$
So $x_n\to 0$, and $\frac{2\pi n}{2n\pi+u_n}\to 1$, so $$\left(1+\sin\left(\frac3{x_n}\right)\right)^{x_n}\to e^{-\alpha}$$
So the above limit cannot exist. It has all of $[0,1]$ as limit points.
Thomas Andrews
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