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By the looks of it, I would say the following is a Neperian limit: $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}$$ but I could not find a way to algebraically bring it in the form: $$\lim_{x\to \infty}\biggl(1+\frac{k}{x}\biggr)^{mx} = e^{mk}$$

Any suggestion on how to solve this?

Johnny Bueti
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5 Answers5

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Hint: $$\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}=\left(\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{1/\sin(2/x^2)}\right)^{2\lim_{x\to \infty}\frac{\sin\frac2{x^2}}{\frac2{x^2}}}$$

Now for the exponent set $\dfrac2{x^2}=h\implies h\to0^+$

lab bhattacharjee
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For questions like this one, I've found asymptotic calculations to be handy. For example:

$$\left(1+\sin\frac{2}{x^2}\right)^{x^2}=e^{x^2\ln\left(1+\sin\frac{2}{x^2}\right)}= e^{x^2\left(\sin\frac{2}{x^2}+o\left(\sin\frac{2}{x^2}\right)\right)}=e^{x^2\left(\frac{2}{x^2}+o\left(\frac{2}{x^2}\right)+o\left(\frac{2}{x^2}\right)\right)}=e^{2+o(1)}\to e^2 (x\to\infty)$$

  • I have never seen anything like this, tried to Google it but did not quite find anything. What is this sorcery? It looks very, very handy, thank you! – Johnny Bueti Feb 14 '18 at 15:44
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    @Fehniix I've had a very quick look and there are some pointers in an earlier Math.SE question: https://math.stackexchange.com/q/1519347/491874. I personally learnt this many years ago and cannot point to a particular source at the moment. –  Feb 14 '18 at 17:12
  • Love it, thank you! Would something of this kind be possible? $$\sin \frac{2}{x^2} \approx \frac{2}{x^2}, \ln \frac{2}{x^2} \approx \frac{2}{x^2}$$ and you would be left with $$\lim_{x \to \infty} e^{x^2\bigl(\frac{2}{x^2}\bigr)} = e^2$$? – Johnny Bueti Feb 15 '18 at 06:26
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    @Fehniik You presumably mean $\ln(1+2/x^2)\sim 2/x^2$... I think the intermediate steps would need justification, because $f(x)\sim g(x)$ doesn't always imply $h(f(x))\sim h(g(x))$ for an arbitrary function $h$. (Example: $x\sim x+\ln x$ when $x\to\infty$, but $e^x\not\sim e^{x+\ln x}=xe^x$...) By using the $o$-notation, you qualify precisely what order of magnitude your error is. –  Feb 15 '18 at 09:30
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Let $${\rm A} = \lim_{x \to a} (f(x))^{g(x)}$$

Here $$\lim_{x \to a} f(x)=1 \quad ; \; \lim_{x \to a} g(x) \to \infty$$

Taking log both the sides

\begin{align} \ln A &=\lim_{x \to a} g(x) \ln (f(x))\\ &=\lim_{x \to a} g(x) \cdot \ln (1+ (f(x)-1)) \\ &=\lim_{x \to a} g(x) \cdot (f(x)-1) \cdot \underbrace{ \lim_{x \to a} \frac{\ln(1+(f(x)-1))}{f(x)-1}}_{= 1} \\ &= \lim_{x \to a} g(x) \times (f(x)-1)\\ \end{align}

Finally we've, this simplification is easy to memorize and is very very helpful. $$\color{blue}{ \lim_{x \to a} (f(x))^{g(x)}= e^{\lim_{x \to a} g(x) \times (f(x)-1)}}$$

Jaideep Khare
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  • Does this work for $a = \infty$? In that case, the expression that wholly evaluates to 1 would evaluate 0, because the denominator grows asymptotically faster than the numerator. The limit itself would then evaluate to zero. Does this apply to this case? – Johnny Bueti Feb 14 '18 at 15:23
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    @Fehniix Yes it does (If $f(x) \to 1$ as $x \to \infty$) . The limit of the expression you are talking about is $1$ because $$\lim_{t \to 0} \frac{\ln (1+t)}{t}=1$$ In that expression $t=f(x)-1$. You get it now? – Jaideep Khare Feb 14 '18 at 17:48
  • That is because $$f(x) = \biggl(1 + \sin \frac{2}{x^2}\biggr)$$ and as $x \to \infty, f(x) \to 1$? What $x$ approaches to by that "special" limit is not required to be $0$? – Johnny Bueti Feb 14 '18 at 20:31
  • @Fehnix I don't understand what exactly is your doubt, can you elaborate? – Jaideep Khare Feb 15 '18 at 05:36
  • I did not quite understand how $$\lim_{t \to 0} \frac{\ln (1+t)}{t}=1$$ worked in relation to $f(x)$ and $g(x)$ in this case, I get it now though, thank you. However, after applying $$\lim_{x \to a} (f(x))^{g(x)}= e^{\lim_{x \to a} g(x) \times (f(x)-1)}$$, I would be left with $$\lim_{x \to \infty} x^2 \cdot \sin \frac{2}{x^2}$$, what would you do at this point? – Johnny Bueti Feb 15 '18 at 08:16
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    @Fehniix Rewrite $$\lim_{x \to \infty} x^2 \cdot \sin \frac{2}{x^2}= \lim_{x \to \infty} \frac{\sin \frac{2}{x^2}}{\frac 1{x^2}} $$Since $x \to \infty$, $\frac{1}{x^2} \to 0$, Hence$$\lim_{x \to \infty} \frac{\sin \frac{2}{x^2}}{\frac 1{x^2}} =\lim_{x \to \infty} \frac{\sin \frac{2}{x^2}}{\frac 2{x^2}} \cdot 2 =2$$ – Jaideep Khare Feb 15 '18 at 08:27
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As an alternative to lab bhattacharjee solution (best approach in my opinion), note that

$$\sin\frac{2}{x^2}=\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)$$

thus

$$\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}=\biggl(1+\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)\biggr)^{x^2}=\left[\biggl(1+\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)\biggr)^{\frac{1}{\frac{2}{x^2}+o\left(\frac{1}{x^2}\right)}}\right]^{2+o(1)}\to e^2$$

user
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You can use this:
If $a \to 1$ and $b \to \infty$, then $\lim a^b = \exp(\lim (a-1)b)$

Botond
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