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From what I learned, $\lim_{x \rightarrow 0} \frac {f(x)}{x^3} = 0$ tells $f(x) = o(x^3)$

In this case, I have tried to compute $\lim_{x \rightarrow 0} \frac {e^{3x^{2}}}{x^3}$, but the limit seems not to exist.

Then I tried to combine these terms to make it available to use L'Hospital's Rule, i.e. $\lim_{x \rightarrow 0} \frac {e^{3x^{2}}}{x^3}-\cos(x^2)$ , $\lim_{x \rightarrow 0} \frac {1}{1+3x^2}-\cos(x^2)$ . However, they all failed.

Any hints or suggestions?

Edit: The definition of little-oh should be $\lim_{x \rightarrow \infty} \frac {f(x)}{g(x)} = 0$ tells $f(x) = o(g)$ rahter than $x \rightarrow 0$

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    $\begingroup$ $f = o(g)$ means that $\lim_{x\to\infty} \frac{f(x)}{g(x)} = 0$ (not the limit as $x\to 0$). $\endgroup$ – rogerl Jan 11 '16 at 1:46
  • $\begingroup$ Can you develop each of the terms as a power series? You only need the first few terms ... $\endgroup$ – Henning Makholm Jan 11 '16 at 1:51
  • $\begingroup$ @rogerl Wiki says $\lim_{x\to\infty} \frac{f(x)}{g(x)} = 0$ holds, but my textbook does teach me to use $\lim_{x \rightarrow 0} \frac {f(x)}{g(x)} = 0$ to prove $f(x) = o(g)$. I think they are equivalent? $\endgroup$ – Jay Wong Jan 11 '16 at 1:55
  • $\begingroup$ They are certainly not equivalent. For example, consider $f(x) = \frac{2}{1+x^2}$ and $g(x) = x+1$. Then $\lim_{x\to 0} \frac{f(x)}{g(x)} = 2$ while $\lim_{x\to\infty} \frac{f(x)}{g(x)}= 0$. $\endgroup$ – rogerl Jan 11 '16 at 1:57
  • $\begingroup$ @rogerl Thank you so much, it seems that my textbook is totally wrong about the definition of little-oh.... $\endgroup$ – Jay Wong Jan 11 '16 at 2:42
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We have $$f(x) = e^{3x^{2}} + \frac{1}{1 + 3x^{2}} - 2\cos(x^{2})$$ and clearly we have \begin{align} L &= \lim_{x \to 0}\frac{f(x)}{x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} + \dfrac{1}{1 + 3x^{2}} - 2}{x^{3}} + 2\cdot\frac{1 - \cos(x^{2})}{x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{(e^{3x^{2}} - 2)(1 + 3x^{2}) + 1}{x^{3}(1 + 3x^{2})} + 2\cdot\frac{1 - \cos(x^{2})}{x^{4}}\cdot x\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} + 3x^{2}e^{3x^{2}} - 6x^{2} - 1}{x^{3}} + 2\cdot\frac{1}{2}\cdot 0\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} - 3x^{2} - 1 + 3x^{2}e^{3x^{2}} - 3x^{2}}{x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} - 3x^{2} - 1}{x^{3}} + 9x\cdot\frac{e^{3x^{2}} - 1}{3x^{2}}\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} - 3x^{2} - 1}{x^{3}} + 9\cdot 0\cdot 1\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} - 3x^{2} - 1}{x^{3}}\notag\\ &= \lim_{x \to 0}\dfrac{e^{3x^{2}} - 3x^{2} - 1}{9x^{4}}\cdot 9x\notag\\ &= 9\lim_{x \to 0}x\cdot\lim_{t \to 0}\frac{e^{t} - t - 1}{t^{2}}\text{ (putting }t = 3x^{2})\notag\\ &= 9\lim_{x \to 0}x\cdot\lim_{t \to 0}\frac{e^{t} - 1}{2t}\text{ (via L'Hospital's Rule)}\notag\\ &= 9 \cdot 0\cdot\frac{1}{2} = 0\notag \end{align} and hence $f(x) = o(x^{3})$ as $x \to 0$.


Update: The notation $o(g(x)), O(g(x))$ is relevant only in case we are trying to describe behavior of some function $f(x)$ in the neighborhood of a point $a$ (or when $x \to \infty$). Thus is is better to mention the point under consideration when using this notation. Thus if $$f(x)/g(x) \to 0 \text{ as } x \to a$$ then we write $$f(x) = o(g(x))\text{ as }x \to a$$ and if $$f(x)/g(x) \to 0\text{ as }x \to \infty$$ then we write $$f(x) = o(g(x)) \text{ as }x \to \infty$$

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  • $\begingroup$ Thank you so much. I am just wondering is it valid that $\lim_{x \rightarrow 0} \frac {f(x)}{x^3} = 0$ tells $f(x) = o(x^3)$ ? Some people here told me we have to use $\lim_{x \rightarrow \infty} \frac {f(x)}{x^3} = 0$? $\endgroup$ – Jay Wong Jan 11 '16 at 5:02
  • $\begingroup$ @JayWong: The notation $o(x)$ assumes that there is some sort of limit of $x$. We should write "$f(x) = o(x^{3})$ as $x \to 0$" but mostly we omit the limit point $0$ or $\infty$ as it is evident from context. I have updated last line of my answer to mention $x \to 0$ explicitly. $\endgroup$ – Paramanand Singh Jan 11 '16 at 5:13
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To see what happens without any presuppositions:

As $x \to 0$,

$\begin{array}\\ e^{3x^{2}} + \frac {1}{1+3x^2} - 2\cos(x^2) &= 1+3x^2+\frac{(3x^2)^2}{2}+O(x^6)+1-3x^2+(3x^2)^2+O(x^6) -2(1-\frac{(x^2)^2}{2}+O(x^8))\\ &= 2+\frac{9x^4}{2}+9x^4+O(x^6) -2+x^4+O(x^8))\\ &= 2+\frac{3(9x^4)}{2}+O(x^6)-2+x^4+O(x^8))\\ &= x^4(\frac{27}{2}+1)+O(x^6)\\ &= x^4(\frac{29}{2})+O(x^6)\\ \end{array} $

Since $x^4 = o(x^3)$, the result is $o(x^3)$.

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If you are attempting this problem using limits and L'Hospital's Rule, keep in mind that $ f(x) = o(g(x)) $ if $$\lim_{x\to\infty} \frac{f(x)}{g(x)} = 0 ,$$ not as x approaches zero. So you need to take the limit as x approaches infinity. In this case, $$ f(x) = e^{3x^2} + \frac{1}{1+3x^2}\ - 2cos(x^2) $$ $$ \text{and} $$ $$ g(x) = x^3. $$

Then you can use L'Hospital's Rule to determine whether or not $f(x)$ is $o(g(x))$.

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  • $\begingroup$ Thank you, I see my mistake. In terms of computing within L'Hospital's Rule, could you please give me some hints? I found it is pretty complex. This question is our previous exam question, so I guess there should be a smart way to deal with it XD $\endgroup$ – Jay Wong Jan 11 '16 at 3:05
  • $\begingroup$ You can possibly avoid using L'Hospital's Rule all together if you observe that f(x) will grow much quicker than g(x), meaning your numerator will grow faster (and be larger) than your denominator. $\endgroup$ – Neil D. Jan 11 '16 at 3:38

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