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$$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $$

I tried writing it as $\lim\limits_{x \to 0} \dfrac{\dfrac{\tan^{12}x}{x^{12}}-1}{x^{2}} $ thinking that if I applied L'hospital two times I would reach my result but with no luck. Any suggestions or tricks I could use?

Paramanand Singh
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The Virtuoso
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3 Answers3

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It is best to first establish the well known limit $$\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{1}{3}\tag{1}$$ This is easily done via the application of L'Hospital's Rule or Taylor series expansions.

The exponent $12$ is used here to intimidate students and one just needs to replace it with a generic symbol $n$. Next we note that for any positive integer $n$ we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\frac{\tan x-x} {x^3}\cdot\sum_{i=1}^{n}\frac{\tan^{i-1}x}{x^{i-1}} =\frac{n}{3}$$ The limit in question is $f(12)=4$.

Thus if one wishes to use L'Hospital's Rule then just a single application of the rule is fine. But one must always use a certain amount of algebraic manipulation before applying the rule.

In case $n$ is not a positive integer then it is best to make use of another standard limit $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{2}$$ along with limit $(1)$. Thus we have $$f(n) =\lim_{x\to 0}\frac{\tan^n x-x^n} {x^{n+2}}=\lim_{x\to 0}\dfrac{\left(\dfrac{\tan x} {x} \right) ^n-1}{\dfrac{\tan x} {x} - 1}\cdot\dfrac{\dfrac{\tan x} {x} - 1}{x^2}\\=\lim_{t\to 1}\frac{t^n-1}{t-1}\cdot\lim_{x\to 0}\frac{\tan x-x} {x^3}=\frac{n}{3}$$ Here we have used the substitution $t=(\tan x) /x$ so that $t\to 1$ as $x\to 0$.

Paramanand Singh
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Hint

The simplest would be to use Taylor series $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$ \frac{\tan^{12}(x)-x^{12}}{x^{14}}=\frac{x^{12}\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -x^{12}}{x^{14} }=\frac 1 {x^2}\left(\left(1+\frac{x^2}{3}+O\left(x^3\right) \right)^{12} -1\right)$$ Now, use the binomial expansion.

Even simpler, remember that, for small $\epsilon$, $(1+\epsilon)^n \sim 1+n\epsilon$

Claude Leibovici
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    Sorry but I don't know how to use Taylor series I need an answer for high school level (if there exists one) – The Virtuoso Jun 14 '18 at 03:50
  • @TheVirtuoso Impress your teacher or prof by using the Taylor series man! Guessing you are a junior or senior...your teacher will likely be very impressed. – Prime Jun 14 '18 at 03:53
  • Junior or senior should know the important power series such as for log, exp, sin, cos, $(1+x)^n, (1-x)^{-n}$. – marty cohen Jun 14 '18 at 04:25
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$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $

$\begin{array}\\ \lim_{x \to 0} \dfrac{\tan^{12}x-x^{12}}{x^{14}} &=\lim_{x \to 0} \dfrac{(\tan(x)/x)^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\sin(x)/(x\cos(x))^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{\sin(x)}{x\cos(x)})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x(1-x^2/2+O(x^4))})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x-x^3/2+O(x^5))})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(\frac{1-x^2/6+O(x^4)}{1-x^2/2+O(x^4)})^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{(1+x^2/3+O(x^4))^{12}-1}{x^{2}}\\ &=\lim_{x \to 0} \dfrac{1+4x^2+O(x^4)-1}{x^{2}}\\ &=\lim_{x \to 0} 4+O(x^2)\\ &=4\\ \end{array} $

marty cohen
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    Is it there no other way besides Taylor? This problem is from an entrance college exam from 2 years ago so there must be one – The Virtuoso Jun 14 '18 at 04:10
  • The problem is that $\sin(x)/x \to 1$ does not have enough precision. Additional terms in sin and cos are needed. – marty cohen Jun 14 '18 at 04:22
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    The question is fundamentally "what is the $x^3$ coefficient of the Maclaurin series of $\tan x$". @TheVirtuoso – Angina Seng Jun 14 '18 at 04:31