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$$ \lim_{x \to 0} \biggl[\frac{\sin x \tan x}{x^2}\biggl] $$where [] is GIF.

The challenge I'm facing is the fact that while $\frac{\sin x}{x}< 1, \frac{\tan x}{x} > 1$. Now their product is becoming a bit undetermined as to will it be $>1$ or $< 1 $. How do i solve it then ??

My first thought was to use the series expansion, which yields $\biggr[1+ \frac{x^2}{3}...\biggl]$ how can I be sure that the further negative terms that are coming in this infinite series will not overpower the $\frac{x^2}{3}$ term and make the whole value < 1

Mr.HiggsBoson
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2 Answers2

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For small values of $x$ use $\sin x \approx x-x^3/6$ and $\tan x \approx x+x^3/3$, then $$\lim_{x\rightarrow 0} \left[ \frac{\sin x \tan x}{x^2} \right]=\lim_{x \rightarrow 0} \left [ (1-x^2/6) (1+x^2/3) \right]=\lim_{x \rightarrow 0} [1+x^2/6-x^4/18]=1.$$

Clayton
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Z Ahmed
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  • This was my first thought, but how can I be sure that the further negative terms that are coming in this infinite series will not overpower the $\frac{x^3}{3}$ term and make the whole value < 1 – Mr.HiggsBoson Jul 29 '19 at 17:13
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    @Colin McLarty If $x^2/6$ is small, $x^4/18$ will be much smaller such that $\frac{x^2}{6}(1-\frac{x}{3})$ is positive if $x$ very small positive/negative. – Z Ahmed Jul 29 '19 at 17:20
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    Actually, the Taylor expansion of the function in brackets is $1+\frac{x^2}{6}+\frac{31x^4}{360}+\cdots$. To get the correct $x^4$ term you would need to take into account the $x^4$ terms of $\frac{\sin x}{x}$ and $\frac{\tan x}{x}$. – Daniel Schepler Jul 29 '19 at 17:36
  • @Daniel Schepler Since the denominator is $x^2$ expanding $\sin x$ and $\cos x$ up to cubic terms is sufficient. – Z Ahmed Jul 29 '19 at 17:46
  • @ZAhmed for $x\to0^+$, I believe the usage of $>$ instead of $\approx$ might be better. Similarly for $x\to0^-$, $<$ – DatBoi Jan 08 '22 at 08:39
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We have that $$\frac{\sin(x)}{x}=1-\frac{x^2}{6}+o(x^2)$$ And $$\frac{\tan(x)}{x}=1+\frac{x^2}{3}+o(x^2)$$ Hence $$\frac{\sin(x)\tan(x)}{x^2}=1+\frac{x^2}{6}+o(x^2)$$ Which means that $\forall \varepsilon > 0 \exists \delta >0$ so that $|x|<\delta \implies |r(x)|<\varepsilon x^2$, where $r(x)$ is the last $o(x^2)$ remainder. This means that $$-\varepsilon x^2 < r(x)<\varepsilon x^2$$ Which means that $$1+\left( \frac{1}{6}+\varepsilon \right) x^2>\frac{\sin(x)\tan(x)}{x^2} > 1+\left( \frac{1}{6}-\varepsilon \right)x^2$$ Which proves that the desired limit is $1$.

Botond
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